Important Questions

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Very Short Answer Type Questions

Question 1.
Write the structures of the products when Butan-2-ol reacts with the following:
(i) CrO₃
(ii) SOCl₂
Answer:

Question 2.
What happens when ethanol reacts with CH₃COCl/pyridine ?
Answer:

Question 3.
When phenol is created with bromine water, while precipitate is obtained. Prove the structure and the name of the compound formed.
Answer:
When phenol is treated with bromine water, white ppt. of 2, 4, 6-tribromophenol is obtained.

Question 4.
Answer the following questions :
(i) Dipole moment of phenol is smaller than that of methanol. Why?
(ii) In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why ?
Answer:
(i) In phenol, C—O bond is less polar due to electron-withdrawing effect of benzene ring whereas in methanol, C—O bond is more polar due to electron-releasing effect of —CH₃ group.

(ii) Phenoxide ion is more reactive than phenol towards electrophilic aromatic substitution and hence undergoes electrophilic substitution with carbon dioxide which is a weak electrophile.

Question 5.
What is denatured alcohol ?
Answer:
Alcohol is made unfit for drinking by mixing some copper sulphate and pyridine in it. This is called denatured alcohol.

Question 6.
Arrange the following compounds in the increasing order of their acidic strength: p-cresol, p -nitrophenol, phenol
Answer:
p-cresol < phenol < p-nitrophenol

Question 7.
Arrange the following compounds in decreasing order of acidity.
(i) H₂O, ROH, HC ☰ CH
(ii)
(iii) CH₃OH, H₂O, C₆H₆OH
Answer:
(i) H₂O > ROH > HC ☰ CH
(ii)
(iii) C₆H₅OH > H₂O > CH₃OH

Question 8.
Suggest a reagent for conversion of ethanol to ethanal.
Answer:
Ethanol can be oxidises into ethanal by using pyridinium chlorochromate.

Question 9.
Explain why sodium metal can be used for drying diethyl ether but not ethyl alcohol.
Answer:
Due to presence of an active hydrogen atom, ethyl alcohol reacts with sodium metal.
2CH₃ — CH₂ — OH + 2Na → 2CH₃ — CH₂ — ONa + H₂
Diethyl ether, on the other hand, does not have replaceable hydrogen atom therefore does not react with sodium metal hence can be dried by metallic sodium.

Question 10.
Phenol is an acid but does not react with sodium bicarbonate solution. Why?
Answer:
Phenol is a weaker acid than carbonic acid (H₂CO₃) and hence does not liberate CO₂from sodium bicarbonate.

Question 11.
In the process of wine making, ripened grapes are crushed so that sugar and enzyme should come in contact with each other and fermentation should start. What will happen if anaerobic conditions are not maintained during this process?
Answer:
Ethanol will be converted into ethanoic acid.

Short Answer Type Questions

Question 1.
Why is the reactivity of all the three classes of alcohols with cone. HCl and ZnCl₂ (Lucas reagent) different ?
Answer:
The reaction of alcohols with Lucas reagent (cone. HCl and ZnCl₂) follow S_N1 mechanism. S_N1 mechanism depends upon the stability of carbocations (intermediate). More stable the intermediate carbocation, more reactive is the alcohol.

Tertiary carbocations are most stable among the three classes of carbocations and the order of the stability of carbocation is 3° > 2° > 1°. This order, intum, reflects the order of reactivity of three classes of alcohols i. e., 3° > 2° > 1°.

Thus , as the stability of carbocations are different so the reactivity of all the three classes of alcohols with Lucas reagent is different.

Question 2.
Write the mechanism of the following reaction:

Answer:
S_N2 mechanism

Question 3.
Explain a process in which a biocatalyst is used in industrial preparation of a compound known to you.
Answer:
Enzymes are biocatalyst. These biocatalysts (enzymes) are used in the industrial preparation of ethanol. Ethanol is prepared by the fermentation of molasses—a dark brown coloured syrup left after crystallisation of sugar which still contains about 40% of sugar.

The process of fermentation actually involves breaking down of large molecules into simple ones in the presence of enzymes. The source of these enzymes is yeast. The various reactions taking place during fermentation of carbohydrates are :

In wine making, grapes are the source of sugars and yeast. As grapes ripen, the quantity of sugar increases and yeast grows on the outer skin. When grapes are crushed, sugar and the enzyme come in contact and fermentation starts. Fermentation takes place in anaerobic conditions i.e., in absence of air. CO₂ gas is released during fermentation.

The action of zymase is inhibited once the percentage of alcohol ,formed exceeds 14 per cent. If air gets into fermentation mixture, the oxygen of air oxidises ethanol to ethanoic acid which in turn destroys the taste of alcoholic drinks.

Question 4.
Explain why alcohols and ethers of comparable molecular mass have different boiling points ?
Answer:
Boiling point depends upon the strength of intermolecular forces of attraction. Higher these forces of attraction, more will be the boiling point. Alcohols undergo intermolecular hydrogen bonding. So, the molecules of alcohols are held together by strong intermolecular forces of attraction.

But in ethers no hydrogen atom is bonded to oxygen. Therefore, ethers are held together by weak dipole-dipole forces, not by strong hydrogen bond.

Since, lesser amount of energy is required than to break weak dipole-dipole forces in ethers than to break strong hydrogen bonds in alcohol.

Question 5.
Explain why is O ＝ C ＝ O non-polar while R—O—R is polar ?
Answer:
CO₂ is a linear molecule. The dipole moment of two C —O bonds are equal and opposite and they cancel each other and hence the dipole moment of CO₂ is zero and it is a non-polar molecule.

While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipoles and hence R—O—R is a polar molecule.

Question 6.
Give reasons for the following:
(i) p-Nitrophenol is more acidic than o-nitrophenol
(ii) Bond angle C—O—C in ethers is slightly higher than the tetrahedral angle (109°28′).
(iii) (CH₃)₃C—Br on reaction with NaOCH₃ gives an alkene instead of an ether.
Answer:
(i)
Intramolecular H-bonding in o-nitrophenol makes loss of proton difficult. Therefore, p-nitrophenol is more acidic than o-nitrophenol.

(ii) The bond angle in ether is slightly higher than 109 °28′ due to repulsive interaction between the two bulky alkyl groups.

(iii) It is because NaOCH₃ is a strong nucleophile as well as a strong base. Thus, elimination reaction predominates over substitution reaction.

Question 7.
Explain the following behaviours :
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol.
(iii) Cumene is a better starting material for the preparation of phenol.
Answer:
(i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses because of H-bond formation between alcohol and water molecules.
(ii) Ortho-nitrophenol is more acidic than ortho-methoxyphenol because nitro being the electron with drawing group stabilises the phenoxids ion.
(iii) Cumene is a better starting material for the preparation of phenol because side product formed in this reaction is acetone which is another important organic compound.

Long Answer Type Questions

Question 1.
(a) Name the starting material used in the industrial preparation of phenol.
(b) Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.
(c) Explain why Lewis acid is not required in bromination of phenol?
Answer:
(a) The starting material used in the industrial preparation of phenol is cumene.

(b) Phenols when treated with bromine water gives polyhalogen derivatives in which all the hydrogen atoms present at ortho and para positions with respect to —OH group are replaced by bromine atoms.

However, in non-aqueous medium such as CS₂, CCl₄, CHCl₃ monobromophenols are obtained.

In aqueous solution, phenol ionises to form phenoxide ion. This ion activates the benzene ring to a very large extent and hence the substitution of halogen takes place at all three positions.

On the other hand, in non-aqueous solution ionisation of phenol is greatly suppressed. Therefore, ring is activated slightly and hence monosubstitution occur.

(c) Lewis acid is an electron deficient molecule. In bromination of benzene, Lewis acid is used-to polarise Br₂ to form Br+ electrophile.

In case of phenol, oxygen atom of phenol itself polarises the bromine molecule to form Br+ ion (electrophile). So, Lewis acid is not required in the bromination of phenol.

Question 2.
Explain the mechanism of the following reactions :
(i) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(ii) Acid catalysed dehydration of an alcohol forming an alkene.
(iii) Acid catalysed hydration of an alkene forming an alcohol.
Answer:
(i) Step I : Nucleophilic addition of Grignard reagent to carbonyl group.

Step II : Formation of carbocation : It is the slowest step and hence, the rate determining step.

To drive the equilibrium to the right, ethylene is removed as it is formed.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Very Short Answer Type Questions

Question 1.
Out of and , which is an example of allylic halide?
X
Answer:
is an example of allylic halide.

Question 2.
Which of the following reactions is S_N1 type ?

Answer:
Reaction (ii) is S_N1 reaction.

Question 3.
Which one of the following compounds is more easily hydrolysed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂CH₂Cl
Answer:
Due to +1 effect of alkyl groups the 2° carbonium ion CH₃—CH—CH₂—CH₃ derived from sec-butyl chloride is more stable than the 1° carbonium ion \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}\) derived from n-propyl chloride. Therefore sec-butyl chloride gets hydrolysed more easily than n-propyl chloride under S_N1 conditions.

Question 4.
What is known as a racemic mixture ? Give an example.
Answer:
equimolar mixture of a pair of enantiomers is called racemic mixture. A racemic mixture is optically inactive due to external compensation.

Question 5.
Consider the three types of replacement of group X by group Y as shown here.

This can result in giving compound (A) or (B) or both. What is the process called if
(A) is the only compound obtained ?
(B) is the only compound obtained ?
(A) and (B) are formed in equal proportions ?
Answer:
(i) Retention
(ii) Inversion
(iii) Racemisation.

Question 6.
What is an asymmetric carbon?
Answer:
A carbon which is attached to four different atoms/groups is called asymmetric carbon. For example, the carbon atom in BrCHClI.

Question 7.
What is plane polarized light?
Answer:
A beam of light which has vibration in only one plane is called plane polarized light.

Question 8.
Why iodoform has appreciable antiseptic property ?
Answer:
Iodoform liberate I₂ when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of I₂ not because of iodoform itself.
(responsible for antiseptic property)

Question 9.
How does the ordinary light differ from the plane polarized light?
Answer:
Ordinary light has oscillations in all the directions perpendicular to the path of propagation whereas plane polarised light has all oscillations in the same plane.

Question 10.
What do you .understand by the term optical activity of compounds?
Answer:
The property of certain compounds to rotate the plane of polarzed light in a characteristic way when it is passed through their solutions is called optical activity of compounds.

Short Answer Type Questions

Question 1.
Give reasons for the following:
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH₃ —X.
Answer:
(i) Since I^– ion is a better leaving group than Br^– ion, hence, CH3I reacts faster than CH₃Br in SN₂ reaction with OH^– ion.

(ii) (±) 2-Butanol is a racemic mixture, i.e., there are two enantiomers in equal proportions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

(iii) In CH₃—X the carbon atom is sp²-hybridised while in halobenzene the carbon atom is sp³-hybridised. The sp²-hybridised carbon is more electronegative due to greater s-character and holds the electron pair of C—X bond more tightly than sp³-hybridised carbon with less s-character. Thus, C—X bond length in CH₃—X is bigger than C—X in halobenzene.

Question 2.
Give reasons for the following:
(i) Haloalkanes easily dissolve in organic solvents.
(ii) Halogen compounds used in industry as solvents are alkyl chlorides rather than bromides and iodides.
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attractions between haloalkanes and organic solvent molecules have much the same strength as ones being broken in the separate haloalkanes and solvent molecules.
(ii) Because alkyl chlorides are more stable and more volatile than bromides and iodides.

Question 3.
Write the mechanism of the following reaction

Answer:
S_N2 mechanism

Question 4.
Which would undergo S_N1 reaction faster in the following pairs and why ?

Answer:
(i)
Tertiary halide reacts faster than primary halide because of greater stability of 3°-carbocation.

(ii)
Because the secondary carbocation formed in the slowest step is more stable than the primary carbocation.

Question 5.
Give reasons:
(i) n-Butyl bromide has higher boiling point than f-butyl bromide. Racemic mixture is optically inactive.
(ii) The presence of nitro group (—NO₂) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Answer:
(i) n-butyl bromide being a straight chain alkyl halide has larger surface area than tert butyl bromide. Larger the surface area, larger the magnitude of the van der Waal’s forces and hence higher is the boiling point.

(ii) A racemic mixture contains the two enantiomers d and l in equal •proportions. As the rotation due to one enantiomer is cancelled by equal and opposite rotation of another enantiomer, therefore, it is optically inactive.

(iii) The presence of NO₂ group at o/p position in haloarenes helps in the stabilisation of resulting carbanion by -R and -I effects and hence increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Question 6.
(i) Why are alkyl halides insoluble in water ?
(ii) Why is butan-l-ol optically inactive but butan-2-ol is optically active ?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why ?
Answer:
(i) This is due to the inability of alkyl halide molecule to form intermolecular hydrogen bonds with water molecules.
(ii)
due to presence of a chiral carbon butan-2-ol is an optically active compound.

(iii) As the weaker resonance (+R) effect of Cl which stabilise the carbocation formed tends to oppose the stronger inductive (-I) effect of Cl which destabilise the carbocation at ortho and para positions and makes deactivation less for ortho and para position.

Question 7.
(i) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change.
Answer:
(i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance.
(ii) Hydride ion (H )

Question 8.
Give the ITJPAC name of the product formed when :
(i) 2-Methyl-l-bromopropane is treated with sodium in the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite.
Answer:
(i) 2, 5-dimethylhexane
(ii) 1 -Methyl-1 -iodocyclohexane
(iii) Nitroethane

Long Answer Type Questions

Question 1.
Some alkyl halides undergo substitution whereas some undergo elimination reaction on treatment with base. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference.
Answer:
Primary alkyl halides follow S_N2 mechanism in which a nucleophile attacks at 180° to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. In S_N2 mechanism, substitution of nucleophile takes place as follows :

Thus, in S_N2 mechanism, substitution takes place. Tertiary alkyl halides follow? S_N1 mechanism, In this case, tert-alkyl halides form 3° carbocation. Now, if the reagent used is a weak base then substitution occur while if it is a strong base then instead of substitution elimination occur.

As alc. KOH is a strong base, so elimination competes over substitution and alkene is formed.

Question 2.
Why are aryl halides less reactive towards nucleophilic substitution reactions than alkyl halides ? How can we enhance the reactivity of aryl halides ?
Answer:
Aryl halides are less reactive towards nucleophilic substitution reaction due to the following reasons :
(i) In haloarenes, the lone pair of electron on halogen are in resonance with benzene ring. So, C—Cl bond acquires partial double bond character which strengthen C—Cl bond. Therefore, they are less reactive towards nucleophilic substitution reaction.

(ii) In haloarenes, the carbon atom attached to halogen is sp²-hybridised. The sp²-hybridised carbon is more electronegative than sp³-hybridised carbon. This sp ²-hybridised carbon in haloarenes can hold the electron pair of C—X bond more tightly and make this C—Cl bond shorter than C—Cl bond haloalkanes.

(iii) Since, it is difficult to break a shorter bond than a longer bond therefore haloarenes are less reactive than haloarenes.
In haloarenes, the phenyl cation will not be stabilised by resonance therefore S_N1 mechanism ruled out.

(iv) Because of the repulsion between the nucleophile and electron rich arenes, aryl halides are less reactive than alkyl halides.

The reactivity of aryl halides can be increased by the presence of an electron withdrawing group (—NO₂) at ortho and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted with OH^– ion.



From the above resonance, it is very clear that electron density is rich at ortho and para positions. So, presence of EWG will facilitate nucleophilic at ortho and para positions not on meta position.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 9 Coordination Compounds Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 9 Coordination Compounds

Very Short Answer Type Questions

Question 1.
Why is CO a stronger ligand than Cl^– ?
Answer:
CO forms π bonds so it is a stronger ligand than Cl^–.

Question 2.
What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?
Answer:
When white light falls on the complex, some part of it is absorbed. Higher the crystal field splitting, lower will be the wavelength absorbed by the complex. The observed colour of complex is the colour generated from the wavelength left over.

Question 3.
How many isomers are there for octahedral complex [CoCl₂ (en) (NH₃)₂]⁺?
Answer:
There will be three isomers: cis and trans isomers. Cis will also show optical isomerism.

Question 4.
Why are low spin tetrahedral complexes not formed?
Answer:
Because for tetrahedral complexes, the crystal field stabilisation energy is lower than pairing energy.

Question 5.
A complex of the type [M(AA)₂X₂]ⁿ⁺ is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
Answer:
An optically active complex of the type [M(AA)₂X₂]ⁿ⁺ indicates cis- octahedral structure, e.g., cis-[Pt(en)₂Cl₂]²⁺ or cis-[Cr(en)₂Cl₂]⁺.

Question 6.
Why is the complex [Co(en)₃]³⁺ more stable than the complex [CoF₆]^3-?
Answer:
Due to chelate effect as the complex [Co(en)₃]³⁺ contains chelating ligand \(\ddot{\mathrm{NH}}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\ddot{\mathrm{NH}}_{2}\).

Question 7.
What do you understand by ‘denticity of a ligand’?
Answer:
The number of coordinating groups present in ligand is called the denticity of ligand. For example, denticity of ethane-1, 2-diamine is 2, as it has two donor nitrogen atoms which can link to central metal atom.

Question 8.
What type of isomerism is shown by the complex [CO(NH₃)₅(SCN)]²⁺?
Answer:
Linkage isomerism.

Question 9.
Arrange the following complex ions in increasing order of crystal field splitting energy △₀ :
[Cr(Cl)₆]^3-, [Cr(CN)₆]^3-, [Cr(NH₃)₆]³⁺
Answer:
[Cr(Cl)₆]^3- < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]^3-

Question 10.
A coordination compound with molecular formula CrCl₃.4H₂O precipitates one mole of AgCl with AgNO₃ solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound?
Answer:
[Cr(H₂O)₄Cl₂] Cl
[Tetraaquadichloridochromium (III) chloride]

Short Answer Type Questions

Question 1.
Give the electronic configuration of the following complexes on the basis of crystal field splitting theory.
[CoF₆]^3-, [Fe(CN)₆]^4- and [Cu(NH₃)₆]²⁺
Answer:
[CoF₆]^3-: Co³⁺(d⁶) \(t_{2 g}^{4} e_{g}^{2}\)
[Fe(CN)₆]^4- : Fe²⁺ (d⁶) \(t_{2 g}^{6} e_{g}^{0}\)
[Cu(NH₃)₆]²⁺ : Cu²⁺ (d⁹) \(t_{2 g}^{6} e_{g}^{3}\)

Question 2.
(i) What type of isomerism is shown by [Co(NH₃) ₅ONO]Cl₂?
(ii) On the basis of crystal field theory, write the electronic configuration for d⁴ ion, if △₀ < P.
(iii) Write the hybridisation and shape of [Fe(CN)₆]^3-.
(Atomic number of Fe = 26)
Answer:
(i) Linkage isomerism and the linkage isomer is [Co(NH₃) ₅ONO]Cl₂.
(ii) If △₀ < P, the fourth electron enters one of two e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\).
(iii) Fe³⁺ : 3d⁵ 4s⁰

Question 3.
Explain why [Fe(H₂O)₆]³⁺ 5.92 BM whereas [Fe(CN)₆]^3- has a value of only 1.74 BM.
Answer:
[Fe(CN)₆]^3- involves d²sp³ hybridisation with one unpaired electron and [Fe(H₂O)₆]³⁺ involves sp³d² hybridisation with five unpaired electrons. This difference is due to the presence of strong CN^– and weak ligand H₂O in these complexes.

Question 4.
CuSO₄∙5H₂O is blue in colour while CuSO₄ is colourless. Why?
Answer:
In CuSO₄∙5H₂O, water acts as ligand as a result it causes crystal field splitting. Hence, d-d transition is possible in CuSO₄∙5H₂O and shows colour. In the anhydrous CuSO₄ due to the absence of water (ligand), crystal field splitting is not possible and hence it is colourless.

Question 5.
Why do compounds having similar geometry have different magnetic moment?
Answer:
It is due to the presence of weak and strong ligands in complexes, if CFSE is high, the complex will show low value of magnetic moment and vice versa, e.g., [CoF₆]^3- and [Co(NH₃)₆]³⁺, the former is paramagnetic and the latter is diamagnetic.

Question 6.
A metal ion Mⁿ⁺ having d⁴ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming △₀ > P:
(i) Write the electronic configuration of d⁴ ion.
(ii) What type of hybridisation will Mⁿ⁺ ion has?
(iii) Name the type of isomerism exhibited by this complex.
Answer:
(i) \(t_{2 g}^{4} e_{g}^{0}\)
(ii) sp³d²
(iii) Optical isomerism

Long Answer Type Questions

Question 1.
Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following: [COF₆]^3-, [CO(H₂O)₆]²⁺, [CO(CN)₆]³
Answer:
Magnetic moment, μ = \(\sqrt{n(n+2)}\)
Where, n = Number of unpaired electrons

No unpaired electrons, so it is diamagnetic.

Question 2.
(i) Draw the geometrical isomers of complex [Pt(NH₃)₂Cl₂].
(ii) Write the hybridisation and magnetic behaviour of the complex [Ni(CO)₄].
(Atomic no. of Ni = 28)
Answer:

Geometrical isomers of [Pt(NH₃)₂Cl₂]

(ii) The complex [Ni(CO)₄] involves sp³ hybridisation.

The complex is diamagnetic as evident from the absence of unpaired electrons.

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 4 Chemical Kinetics

Very Short Answer Type Questions

Question 1.
For which type of reactions, order and molecularity have the same value?
Answer:
If the reaction is an elementary reaction, order is same as molecularity.

Question 2.
Why is the probability of reaction with molecularity higher than three very rare?
Answer:
The probability of more than three molecules colliding simultaneously is very small. Hence, possibility of molecularity being three is very low.

Question 3.
State a condition under which a bimolecular reaction is kinetically first order.
Answer:
A bimolecular reaction may become kinetically of first order if one of the reactants is in excess.

Question 4.
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
Answer:
Thermodynamically the conversion of diamond to graphite is highly feasible but this reaction is very slow because its activation energy is high.

Question 5.
Why is it that instantaneous rate of reaction does not change when a part of the reacting solution is taken out?
Answer:
Instantaneous rate is measured over a very small interval of time, hence, it does not change when a part of solution is taken out.

Question 6.
A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction?
Solution:
As t_75% = 2t_50%
Therefore, it is a first order reaction.

Question 7.
Define threshold energy of a reaction.
Answer:
Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecules.

Question 8.
Why does the rate of a reaction increase with rise in temperature?
Answer:
At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy), which leads to faster rate.

Question 9.
What is the difference between rate law and law of mass action?
Answer:
Rate law is an experimental law. On the other hand, law of mass action is a theoretical law based on the balanced chemical reaction.

Question 10.
What do you understand by ‘Rate of reaction’?
Answer:
The change in the concentration of any one of the reactants or products per unit time is termed as the rate of reaction.

Question 11.
In the Arrhenius equation, what does the factor e a corresponds to?
Answer:
e^-E_a/RT corresponds to the fraction of molecules that have kinetic energy greater than E_a,

Short Answer Type Questions

Question 1
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L^-1mols^-1
(ii) Lmol^-1s^-1.
Solution:
The sum of powers of the concentration of the reactants in the rate law expression is called order of reaction.
For a general reaction: aA + bB → Products
If rate = k[A]^m [B]ⁿ; order of reaction = m + n

(i) General unit of rate constant, k = (mol L^-1 )^1-ns^-1
L^-1mol s^-1 = (mol L^-1 )^1-ns^-1
-1 = -1 + n ⇒ n = 0 ∴ Reaction order = 0

(ii) L mol^-1 s^-1 = (mol L^-1)^1-n s^-1
1 = -1 + n ⇒ n = 2 ∴ Reaction order = 2

Question 2.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation :
log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\)K
Calculate E_a for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 JK^-1 mol^-1)
Solution:
Comparing the equation, log k = 14.2 – \(\frac{1.0 \times 10^{4}}{T}\) K with the equation,
log k = log A = \(\frac{E_{a}}{2.303 R T}\), we get
\(\frac{E_{a}}{2.303 R}\) = 1.0 × 10⁴ K or E_a = 1.0 × 10⁴ K × 2.303 × R
E_a = 1.0 × 10⁴ K × 2.303 × 8.314 JK^-1
= 19.1471 × 10⁴ Jmol^-1
= 191.47 kJ mol^-1
For a first order reaction, t_t/2 = \(\frac{0.693}{k}\) or k = \(\frac{0.693}{t_{1 / 2}}\)
k = \(\frac{0.693}{200 \mathrm{~min}}\) = 3.465 × 10 ^-3min^-1

Question 3.
The reaction, N₂(g) + O₂(g) ⇌ 2NO(g) contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 × 10^-5. Suppose in a case [N₂] = 0.80 mol L^-1 and [O₂] = 0.20 mol L^-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
Solution:

[N₂] = 0.8 – 6.324 × 10⁴ mol L^-1
= 0.799 molL^-1
[O₂] = 0.2 – 6.324 × 10^-4 mol L^-1
= 0.199 mol L^-1

Question 4.
For a general reaction, A → B, plot of concentration of A vs time is given in figure. Answer the following questions on the basis of this graph.

(i) What is the order of the reaction?
(ii) What is the slope of the curve?
(iii) What are the units of rate constant?
Answer:
(i) Zero order
(ii) Slope = – k
(iii) Units of rate constant = mol L^-1 s^-1

Question 5.
For a reaction, A + B → products, the rate law is rate = k [A][B]a^3/2. Can the reaction be an elementary reaction? Explain.
Answer:
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is \(\frac{3}{2}\). This indicated that the reaction is not an elementary reaction.

Question 6.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce 1 g of the reactant to 0.0625 g?
Answer:
We know that, t = \(\frac{2.303}{k}\) log \(\frac{[R]_{0}}{[R]}\)
t = \(\frac{2.303}{60}\) log \(\frac{1}{0.0625}\)
t = 0.0462 s

Long Answer Type Questions

Question 1.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	30	60
[CH3COOCH3]/mol L-1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
Solution:

As the value of k is same in both the cases, therefore, hydrolysis of methylacetate in aqueous solution follows pseudo first order reaction.

(ii) Average rate = \(-\frac{\Delta\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]}{\Delta t}\)
= \(\frac{-[0.15-0.30]}{60-30}\) = \(\frac{0.15}{30}\)
Average rate = 0.005 mol L^-1s^-1

Question 2.
Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?
Answer:
A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical change.
According to “intermediate complex formation theory” reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.

The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.

Thus, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy.

We know that, lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.

Enthalpy, △H is a ‘state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram.
Potential energy diagram of catalysed reaction is given as:

Question 3.
All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Answer:
Only effective collision lead to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species).

And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones i.e., in products.
e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply back and no products are formed.

To account for effective collisions, another factor P (probability of steric factor) is introduced K = PZ_ABe^-Ea/RT.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 7 Alternating Current Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 7 Alternating Current

Very short answer type questions

Question 1.
Why is the use of AC voltage preferred over DC voltage? Give two reasons.
Answer:
The use of AC voltage is preferred over DC voltage because of

• the loss of energy in transmitting the AC voltage over long distance with the help of step-up transformers is negligible as compared to DC voltage.
• AC voltage can be stepped up and stepped down as per the requirement by using a transformer.

Question 2.
Explain why current flows through an ideal capacitor when it is connected to an AC source, but not when it is connected to a DC source in a steady state.
Answer:
For AC source, circuit is complete due to the presence of displacement current in the capacitor. For steady DC, there is no displacement current, therefore, circuit is not complete.
Mathematically, capacitive reactance
X_C = \(\frac{1}{2 \pi f C}=\frac{1}{\omega C}\)
So, capacitor allows easy path for AC source.
For DC, / = 0, so X_C = infinity.
So, capacitor blocks DC.

Question 3.
Define capacitor reactance. Write its SI units.
Answer:
Capacitor reactance is the resistance offered by a capacitor, when it is connected to an electric circuit. It is given by X_C = \(\frac{1}{\omega C}\)
where, ω = angular frequency of the source
C = capacitance of the capacitor
The SI unit of capacitor reactance is ohm (Ω).

Question 4.
In a series LCR circuit, V_L = V_C ≠ V_R What is the value of power factor for this circuit?
Answer:
Power factor = 1
Since V_L = V_C, the inductor and capacitor will nullify the effect of each other and it will be a resistive circuit.
For Φ =0; power factor cosΦ = 1

Question 5.
The power factor of an AC circuit is 0.5. What is the phase difference between voltage and current in this circuit?
Answer:
Power factor between voltage and current is given by cosΦ, where Φ is phase difference
cosΦ = 0.5 = \(\frac{1}{2}\) ⇒ Φ = cos^-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}\)

Question 6.
What is wattless current?
Answer:
When pure inductor and/or pure capacitor is connected to AC source, the current flows in the circuit, but with no power loss; the phase difference between voltage and current is \(\frac{\pi}{3}\) . Such a current is called the wattless current.

Question 7.
An AC source of voltage V = V₀ sin ωt is connected to an ideal inductor. Draw graphs of voltage V and current I versus cat.
Answer:
Graphs of V and I versus ωt for this circuit is shown below:

Question 8.
Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit?
Answer:
The quality factor (Q) of series LCR circuit is defined as the ratio of the resonant frequency to frequency band width of the resonant curve.
Q = \(\frac{\omega_{r}}{\omega_{2}-\omega_{1}}=\frac{\omega_{r} L}{R}\)

Clearly, smaller the value of R, larger is the quality factor and sharper the resonance. Thus, quality factor determines the nature of sharpness of resonance. It has no units.

Question 9.
What is the function of a step-up transformer?
Answer:
Step-up transformer converts low alternating voltage into high alternating voltage and high alternating current into low alternating current. The secondary coil of step-up transformer has greater number of turns than the primary (N_s > N_p ).

Question 10.
Mention the two important properties of the material suitable for making core of a transformer.
Answer:
Two characteristic properties:

1. Low hysteresis loss
2. Low coercivity

Question 11.
If an LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy? (NCERTExemplar)
Answer:
Magnetic energy analogous to kinetic energy and electrical energy analogous to potential energy.

Question 12.
A device ‘X’ is connected to an a.c. source. The variation of voltage, current and power in one complete cycle is shown in the figure.

(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’. (NCERT Exemplar)
Answer:
(a) A
(b) Zero
(c) L or C or LC

Short answer type questions

Question 1.
Prove that an ideal capacitor in an AC circuit does not dissipate power.
Answer:
Since, average power consumption in an AC circuit is given by
P_av = V_rms × I_rms × cosΦ
But in pure capacitive circuit, phase difference between voltage and current is given by
Φ = \(\frac{\pi}{2}\)
∴ P_av = V_rms × I_rms × cos \(\frac{\pi}{2}\)
⇒ P_av = 0 ( ∵ cos \(\frac{\pi}{2}\) = 0)
Thus, no power is consumed in pure capacitive AC circuit.

Question 2.
A circuit is set up by connecting inductance L 100 mil, resistor R -100 D. and a capacitor of reactance 200 Ω in series. An alternating emf of 150 √2 V, 500/ π Hz is applied across this series combination. Calculate the power dissipated in the resistor.
Answer:
Here, L =100 x 10^-3 H,R =100 Ω,
X_C = 200 Ω,V_rms = 150√2 V
v = \(\frac{500}{\pi}\) HZ

= 225 W

Question 3.
A series L-C-R circuit is connected to an AC source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation.
Answer:
Assuming X_L > X_C
⇒ V_L > V_C
∵ Net voltage, V = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)
where, V_L, V_C and are alternating voltages across L,C and R respectively.


But, V_R= IR,V_L = IX_L,
V_C = IX_C
∴ Net voltage, V = \(\sqrt{(I R)^{2}+\left(I X_{L}-I X_{C}\right)^{2}}\)
\(\frac{V}{I}\) = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
Impedance of LCR circuit,
Z = \(\frac{V}{I}\) = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

Question 4.
In a series LCR circuit connected to an AC source of variable frequency and voltage V = V_m sin ωt, draw a graph showing the variation of current (I) with angular frequency (ω) for two different values of resistance R₁ and R₂ (R₁ > R₂ ). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance.
Answer:
Figure shows the variation of i_m with ω in a LCR series circuit for two values of resistance R₁ and R₂ (R₁ > R₂).

The condition for resonance in the LCR circuit is
ω₀ = \(\frac{1}{\sqrt{L C}}\)
We see that the current amplitude is maximum at the resonant frequencyω. Since i_m = v_m / R at resonance, the current amplitude for case R₂ is sharper to that for case R₁.

Quality factor or simply the Q-factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.
It is given by Q = \(\frac{1}{R} \cdot \sqrt{\frac{L}{C}}\)
The Q-factor determines the sharpness of the resonance curve. Less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit.

Question 5.
Both alternating current and direct current y, are measured in amperes. But how is the ampere defined for an alternating current? (NCERT Exemplar)

Answer:
An ac current changes direction with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms vc of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of ac.

Question 6.
Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency. (NCERT Exemplar)
Answer:
A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i. e., if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency, it is given byl/©C.

Question 7.
Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
(NCERT Exemplar)
Answer:
An inductor opposes flow of current through it by developing an induced emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e., if the frequency is higher. The reactance of an inductor, therefore, is proportional to the frequency, being given by ωL.

Long answer type questions

Question 1.
(a) An AC source of voltage V = V₀ sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expression for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(b) In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate \(\frac{P_{1}}{P_{2}}\).
Answer:
(a) Expression for Impedance in LCR Series Circuit : Suppose resistance R, inductance L and capacitance C are connected in series and an alternating source of voltage V = V₀ sin ωt is applied across it. (fig. a) On account of being in series, the current (i) flowing through all of them is the same.

Suppose, the voltage across resistance R isV_R, voltage across inductance L is V_L and voltage across capacitance C is V_C. The voltage V_R and current i are in the same phase, the voltage V_L will lead the current by angle 90° while the voltage V_C will lag behind the current by angle 90° (fig. b). Clearly,V_C and V_L are in opposite directions, therefore their resultant potential difference = V_C – V_L (if V_C >,V_L).

Thus, V_R and (V_C – V_L) are mutually perpendicular and the phase difference between them is 90°. As applied voltage across the circuit is V, the resultant of VR and (V_C – V_L) will also be V. From fig.

Question 2.
(i) What do you understand by sharpness of resonance in a series L-C-R circuit? Derive an expression for Q-factor of the circuit.
Three electrical circuits having AC sources of variable frequency are shown in the figures. Initially, the current flowing in each of these is same. If the frequency of the applied AC source is increased, how will the current flowing in these circuits be affected? Give the reason for your answer.

Answer:

The sharpness of resonance in series LCR circuit refers how quick fall of alternating current in circuit takes place when frequency of alternating voltage shifts away from resonant frequency. It is measured by quality factor (Q-factor) of circuit.

The Q-factor of series resonant circuit is defined as the ratio of the voltage developed across the capacitance or inductance at resonance to the impressed voltage which is the voltage applied.

This is the required expression.

(ii) Let initially I_r current is flowing in all the three circuits. If frequency of applied AC source is increased, then the change in current will occur in the following manner.
(a) Circuit Containing Resistance R Only: There will not be any effect in the current on changing the frequency of AC source.

where,f_i = initial frequency of AC source.
There is no effect on current with the increase in frequency.

(b) AC Circuit Containing Inductance

Only: With the increase of frequency current of AC source inductive reactance increase as
I = \(\frac{V_{r m s}}{X_{L}}=\frac{V_{r m s}}{2 \pi f L}\)
For given circuit,
I ∝ \(\frac{1}{f}\)
Current decreases with the increase of frequency.

(c) AC Circuit Containing Capacitor Only:
X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
Current, I = \(\frac{V_{r m s}}{X_{C}}\) = \(\frac{V_{r m s}}{\left(\frac{1}{2 \pi f C}\right)}\)
I = 2πfCV_rms
For given circuit, I ∝ f
Current increases with the increase of frequency.

Question 3.
(a) Describe briefly, with the help of a labelled diagram, the working of a step up transformer.
(b) Write any two sources of energy loss in a transformer.
(c) A step up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain
Or Draw a labelled diagram of a step-down transformer. State the principle of its working. Express the turn ratio in terms of voltages.
Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer.
How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V-550 W refrigerator?
Answer:
(a) Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. It is based on the principle of mutual-induction.

Construction: It consists of laminated core of soft iron, on which two coils of insulated copper wire are separately wound. These coils are kept insulated from each other and from the iron-core, but are coupled through mutual induction. The number of turns in these coils are different. Out of these coils one coil is called primary coil and other is called the secondary coil. The terminals of primary coils are connected to AC mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. Transformers are of two types:
1. Step-up transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil. (i. e.,N_s> N_p).
2. Step-down transformer: It transforms the alternating high voltage to alternating low voltage and in this the number of turns in secondary coil is less than that in primary coil (i. e.N_s < N_p)

Working: When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil; due to which the magnetic flux linked with the secondary coil changes continuously, therefore the alternating emf of same frequency is developed across the secondary.

Let N_p be the number of turns in primary coil, N_s the number of turns in secondary coil and Φ the magnetic flux linked with each turn. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. According to Faraday’s laws the emf induced in the primary coil
ε ₀ = -N_p\(\frac{\Delta \phi}{\Delta t}\) …………….. (1)
and emf induced in the secondary coil
ε _s = -N_p\(\frac{\Delta \phi}{\Delta t}\) ……………… (2)
From eq. (1) and eq, (2)
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) …………………. (3)
If the resistance of primary coil is negligible, the emf (ε _p) induced in the primary coil, will be equal to the applied potential difference (V_p) across its ends. Similarly if the secondary circuit is open, then the potential difference Vs across its ends will be equal to the emf (ε _s) induced in it; therefore,
\(\frac{V_{s}}{V_{p}}=\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) r(say) …………… (4)
where r = \(\frac{N_{S}}{N_{P}}\) is called the transformation ratio. If i_p and i_s are the instantaneous currents in primary and secondary coils and there is no loss of energy.
For about 100% efficiency,
Power in primary = Power in secondary
V_p i_p = V_si_s
∴ \(\frac{i_{s}}{i_{p}}=\frac{V_{p}^{F}}{V_{s}}=\frac{N_{p}}{N_{s}}=\frac{1}{r}\) ………….. (5)

In step-up transformer, N_s > N_p → r > 1 ;
So V_s > V_p and i_s < i_p
i.e., Step up transformer increases the voltage.

In step down transformer, N_s < N_p → r < 1
So V_s < V_p and i_s > i_p
i.e., step-up down transformer decreases the voltage, but increase the current.

Laminated Core: The core of a transformer is laminated to reduce the energy losses due to eddy currents. So, that its efficiency may remain nearly 100%.
In a transformer with 100% efficiency (say), Input power = output power dV_pI_p = V_sI_s
(b) The sources of energy loss in a transformer are, (i) eddy current losses due to iron core, (ii) flux leakage losses, (iii) copper losses due to heating up of copper wires, (iv) Hysteresis losses due to magnetisation and demagnetisation of core.

(c) When output voltage increases, the output current automatically decreases to keep the power same. Thus, there is no violation of conservation of energy in a step-up transformer.
We have, i_p V_p = i_sV_s = 550 W
V_p 220V
i_p = \(\frac{550}{220}=\frac{5}{2}\) = 2.5A

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments

Very short answer type questions

Question 1.
What is reflection?
Answer:
When a light ray incident on a smooth surface bounces back to the same medium, it is called reflection.

Question 2.
State new cartesian sign conventions used for mirrors.
Answer:

• All the distances are measured from the pole of the mirror.
• All the distances measured in the direction of incident ray are taken as positive and the distances measured opposite to the incident ray are taken as – ve.
• All heights measured perpendicular to the principal axis in the upward direction are taken as + ve and those measured in downward direction are taken as – ve.

Note: Direction of incident light is always to be shown falling from left to right. So distance of the object and real image is always -ve while that of virtual image is always + ve, height of real image is always – ve while that of the virtual image and the size of real object are always + ve.

Question 3.
How does focal length of a lens change when red light incident on it is replaced by violet light? Give reason for your answer.
Answer:
The refractive index of the material of a lens increases with the decrease in wavelength of the incident light. So, focal length will decrease with a decrease in wavelength according to the formula.
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Thus, when we replace red light with violet light then due to increase in wavelength the focal length of the lens will decrease.

Question 4.
Define refraction of light.
Answer:
It is defined as the process of bending of light from its path when it travels from one medium to the another.

Question 5.
State
(a) Laws of reflection.
(b) Laws of refraction.
Answer:
(a) The following are the two laws of reflection :
(i) Angle of incidence is always equal to the angle of reflection.
(ii) The incident ray, reflected ray and normal to the surface at the point of incidence all lie in the same plane.

(b) The following are the two laws of refraction :
(i) The ratio of the sine of angle of incidence to the sine of the angle of refraction is always constant for a given pair of media.
i.e., \(\frac{\sin i}{\sin r}\) = constant = ^aµb
where ^aµb is called relative refractive index of medium b w.r.t. a.
(ii) The incident ray, refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

Question 6.
(i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.
Answer:
(i) Refractive index (µ) = \(\frac{1}{\sin C}\)
where, C is the critical angle.
(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths (colours) of light.

Question 7.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer:
A biconvex lens will act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction as itself. In this case, the focal length 1/f = 0 or f→ ∞.

Question 8.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging lens? Give reason.
Answer:
No, it will behave as a diverging lens.
On Using thin lens maker formula
\(\frac{1}{f_{w}}=\left(\frac{n_{g}}{n_{m}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
On Using sign convention R₁ = +ve, R₂ = -ve and n_g = 1.25 and n_m = 1.33
\(\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\) +ve,value and \(\left(\frac{1.25}{1.33}-1\right)\) =-ve value Hence f_w = -ve , so it behaves as a diverging lens.

Question 9.
Define total internal reflection.
Answer:
It is defined as the process of reflection of light that takes place when a ray of light travelling from denser to rarer medium gets incident at the interface of the two media at an angle greater than the critical angle for the given air of media.

Question 10.
State the criteria for the phenomenon of total internal reflection of light to take place.
Answer:
Following are the criteria for total internal reflection

• Light must pass from a denser to a rarer medium.
• Angle of incidence must be greater than critical angle.

Question 11.
Define mirage.
Answer:
It is defined as an optical illusion that occurs in deserts and coal tarred roads appear to be covered with water but on approaching at that place no water is obtained. In deserts thirsty animals observe virtual images of trees on hot sand so expecting a pond of water there but on reaching there, they do not get water pond and hence called optical illusion.

Question 12.
Why diamond sparkles?
Answer:
The critical angle for diamond is low i.e., 23° and its refractive index is 2.47. The faces of diamond are cut in such a way that when a ray of light entering from a face undergoes multiple total internal reflections from its different faces. Due to small value of the critical angle, almost all light rays entering the diamond suffer multiple total internal reflection and thus it shines brilliantly.

Question 13.
What are optical fibres? Give their one use.
Answer:
Optical fibres are thousands of very fine quality fibres of glass or quartz. The diameter of each fibre is of the order of 10^-4 cm having refractive index of material equal to 1.7. These are coated with a thin layer of material having µ = 1.5.
They are used in transmission and reception of electrical signals by converting them first into light signals.

Question 14.
Write the relationship between angle of incidence ‘i’ angle of prism ‘A’ and angle of minimum deviation for a triangular prism.
Answer:
i = \(\frac{A+\delta_{m}}{2}\)
where, δ_m = angle of minimum deviation.

Question 15.
Define dispersion of light. What is its cause?
Answer:
It is defined as the process of splitting up of white light into its constituent colours on passing through a prism.
We know that for small angled prism,
δ = (µ -1)A.
Also according to Cauchy’s formula, we know that µ ∝ \(\frac{1}{\lambda^{2}}\)
Thus µ of the material of prism is different for different colours, so δ is also different for different incident colours.
Thus due to different values of angle of deviation, each colour occupies different direction in emergent beam of light and thus constituent colours of white light get dispersed. λv < λr, so δv > δr.
The violet colour deviates more than the red colour.

Question 16.
Explain the rainbow.
Answer:
The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near western horizon) while it is raining in the opposite part of the sky (say eastern horizon). An observer can therefore see a rainbow only when his back is towards the sun.

Question 17.
Why does the Sun look reddish at sunset or sunrise?
Answer:
During sunset or sunrise, the sun is just above the horizon, the blue colour gets scattered most by the atmospheric molecules while red light gets scattered least, hence Sun appears red.
I ∝ \(\frac{1}{2^{4}}\) and λ_B << λ_R.

Question 18.
Will the focal length of a lens for red light be more, same or less than that for blue light? (NCERT Exemplar)
Answer:
As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red. Thus the focal length for red light will be more than that for blue.

Question 19.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed? (NCERT Exemplar)
Answer:
No, the reversibility of the tens makes equation.
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= -(n-1) \(\left(\frac{1}{R_{2}}-\frac{1}{R_{1}}\right)\)
On reversing the lens, values of R₁ and R₂ are reversed and so their signs.
Hence, for a given position of object (u), position of image (v) remains unaffected.

Question 20.
Why danger signals are of red light?
Answer:
Scattering of light is inversely proportional to the fourth power of wavelength of incident light. As red light has longer wavelength as compared to other visible colours, so its scattering is least and thus red light signals can be seen from a longer distance.

Short answer type questions

Question 1.
Will the focal length of a lens for red light be more, same or less than that for blue light? [NCERT Exemplar]
Answer:
As the refractive index for red is less than that for blue parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
In other words, μ_b > μ_r By lens maker’s formula,
\(\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Therefore, f_b < f_r
Thus, the focal length for blue light will be smaller than that for red.

Question 2.
Define power of a lens. Write its units. Deduce the relation \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) for two thin lenses kept in contact coaxially.
Answer:
The power of a lens is equal to the reciprocal of its focal length when it is measured in metre. Power of a lens,
P = \(\frac{1}{f(\text { metre })}\)
Its SI unit is dioptre (D).

Consider two lenses A and B of focal lengths, f₁ and f₂ placed in contact with each other. An object is placed at a point O beyond the focus of the first lens A. .
The first lens produces an image (real image) at I₁ which serves as a virtual object for the second lens B producing the final image at I.

Since, the lenses are thin, we assume the optical centres P of the lenses to be coincident. For the image formed by the first lens A, we obtain
\(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) ……………………………. (1)
For the image formed by the second lens B, we obtain
\(\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}\) …………………………………. (2)
Adding eqs. (1) and (2), we obtain
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …………………………. (3)

If the two lenses system is regarded as equivalent to a single lens of focal length f, we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ……………………………… (4)
From eqs. (3) and (4), we obtain
\(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{f}\) .

Question 3.
(a) Draw a schematic labelled ray diagram of a reflecting type telescope (cassegrain).
(b) The objective of telescope is of larger focal length and of larger aperture (compared to the eyepiece). Why? Given reasons.
(c) State the advantages of reflecting telescope over refracting telescope.
Answer:
(a)

(b) In normal adjustment, magnifying power of the telescope, M = \(\frac{f_{0}}{f_{e}}\)
(i) If focal length of the objective lens is large in comparison to the eyepiece, magnifying power increases.
(ii) Resolving power of the telescope RP = \(\frac{D}{1.22 \lambda}\)
D being the diameter of the objective. To increase the resolving power of the telescope, large aperture of the objective lens is required.

Advantages

• There is no chromatic aberration in a mirror.
• Brighter image.
• High resolving power.
• Large magnifying power.

Question 4.
How is the working of a telescope different from that of a microscope?
Answer:
Difference in working of telescope and microscope

• Objective of telescope forms the image of a very far off object at or within the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
• The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size.
• The objective of a telescope has large focal length and large aperture while the corresponding parameters for a microscope have very small ‘ values.

Question 5.
For a glass prism (μ = \(\sqrt{3}\) ) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. (NCERT Exemplar)
Answer:
At minimum deviation μ = \(\frac{\sin \left[\frac{\left(A+\delta_{m}\right)}{2}\right]}{\sin \left(\frac{A}{2}\right)} \)
Given, δ_m = A
∴ μ = \(\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}\)
∴ \(\cos \frac{A}{2}=\frac{\sqrt{3}}{2} \text { or } \frac{A}{2}=30\)
⇒ A = 60⁰.

Long Answer Type Questions

Question 1.
(a) Draw a ray diagram for formation of image of a point object by a thin double convex lens having radii of curvature R₁ and R₂. Hence, derive lens maker’s formula for a double convex lens. State the assumptions made and sign convention used.
(b) A convex lens is placed over a plane mirror. A pin is now positioned so that there is no parallax between the pin and its image formed by this lens-mirror combination. How will you use this observation to find focal length of the lens? Explain briefly.
Answer:
(a) Lens Maker’s Formula: Suppose L is a thin lens. The refractive index of the material of lens is n₂ and it is placed in a medium of refractive index n₁. The optical centre of lens is C and X’ X is principal axis. The radii of curvature of the surfaces of the lens are R₁ and R₂ and their poles are P₁ and P₂.

The thickness of lens is t, which is very small. O is a point object on the principal axis of the lens. The distance of O from pole P₁ is u. The first refracting surface forms the image of O at I’ at a distance v’ from P₁.
From the refraction formula at spherical surface, \(\frac{n_{2}}{v^{\prime}}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R_{1}}\) ……………………………….. (1)
The image I’ acts as a virtual object for second surface and after refraction at second surface, the final image is formed at I.

The distance of I from pole P₂ of second surface is v. The distance of virtual object (I’) from pole P₂ is (v’ -t).
For refraction at second surface, the ray is going from second medium (refractive index n₂) to first medium (refractive index n₁), therefore, from refraction formula at spherical surface

\(\frac{n_{1}}{v}-\frac{n_{2}}{\left(v^{\prime}-t\right)}=\frac{n_{1}-n_{2}}{R_{2}}\) ……………….. (2)
For a thin lens t is negligible as compared to v’, therefore from eq. (2)
\(\frac{n_{1}}{v}-\frac{n_{2}}{v^{\prime}}=-\frac{n_{2}-n_{1}}{R_{2}}\) ……………………………….. (3)
Adding equation (1) and (3),we get

where, ₁n₂ = \(\frac{n_{2}}{n_{1}}\) is refractive index of second medium (te. medium of lens) with respect to first medium. If the object O is at infinity, the image will be formed at second focus i.e., if u = ∞, v = f₂ =f
Therefore, from equation (4)

This formula is called Lens-Maker’s formula. If first medium is air and refractive index of material of lens be n, then ₁n₂ = n, therefore, the modified equation (5) may be written as
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ………………………………. (6)

(b) Focal length = distance of the pin from the mirror.
The rays from the object after refraction from lens should fall normally on the plane mirror. So, they retrace their path. Hence, rays must be originating from focus and thus distance of the pin from the plane mirror gives focal length of the lens.

Question 2.
(a) Draw the labelled ray diagram for the formation of image by a compound microscope. Derive an expression for its total magnification (or magnifying power), when the final image is formed at the near point. Why both objective and eyepiece of a compound microscope must have short focal lengths?
(b) Draw a ray diagram showing the image formation by a compound microscope. Hence, obtain expression for total magnification when the image is formed at infinity.
Answer:
(a) Compound Microscope: It consists of a long cylindrical tube, containing at one end a convex lens of small aperture and small focal length. This is called the objective lens (0). At the other end of the tube another co-axial smaller and wide tube is fitted, which carries a convex lens (E) at its outer end. This lens is towards the eye and is called the eyepiece. The focal length and aperture of eyepiece are somewhat larger than those of objective lens. Cross-wires are mounted at a definite distance before the eyepiece. The entire tube can be moved forward and backward by the rack and pinion arrangement.

Adjustment: First of all the eyepiece is displaced backward and forward to focus it on cross-wires. Now the object is placed just in front of the objective lens and the entire tube is moved by rack and pinion arrangement until there is no parallax between image of object and cross wire. In this position, the image of the object appears quite distinct.

Working: Suppose a small object AB is placed slightly away from the first focus F_o‘of the objective lens. The objective lens forms the real, inverted and magnified image A’ B’, which acts as an object for eyepiece. The eyepiece is so adjusted that the image A’B’ lies between the first focus F_e‘ and the eyepiece E. The eyepiece forms its image A”B” which is virtual, erect and magnified. Thus the final image A”B” formed by the microscope is inverted and magnified and its position is outside the objective and eyepiece towards objective lens.

The magnifying power of a microscope is defined as the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e., Magnifying power,
M = \(\frac{\beta}{\alpha}\)
As object is very small, angles a and 1 are very small and so tan α = α and tan β = β. By definition the object AB is placed at the least distance of distinct vision.

∴ α = tan α = \(\frac{A B}{E A}\)
By sign convention, EA = – D,
∴ α = \(\frac{A B}{-D}\)
and from figure
β = tan β = \(\frac{A^{\prime} B^{\prime}}{E A^{\prime}}\)
If u_e is distance of image A’ B’ from eyepiece E, then by sign convention, EA’ = -u_e
and so, β = \(\frac{A^{\prime} B^{\prime}}{-u_{e}}\)

Hence, magnifying power,
M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} /\left(-u_{e}\right)}{A B /(-D)}=\frac{A^{\prime} B^{\prime}}{A B} \cdot \frac{D}{u_{e}}\)
By sign conventions, magnification of objective lens
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{v_{o}}{\left(-u_{o}\right)}\)
∴ M = \(-\frac{v_{o}}{u_{o}} \cdot \frac{D}{u_{e}}\) ………………………………….. (2)

Using lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) for eyelens,
(i.e. using f = f_e’ V = -v_e, U = -u_e ) we get
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or \(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Magnifying power,
M = \(-\frac{v_{o}}{u_{o}} D\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\)

or M = \(-\frac{v_{o}}{u_{o}}\left(\frac{D}{f_{e}}+\frac{D}{v_{e}}\right)\)
When final image is formed at the distance of distinct vision, V_e = D
∴ Magnification,
M= – \(\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)\)

For greater magnification of a compound microscope, f_e should be small. As f_o < f_e’ so f₀ is small.
Hence, for greater magnification both f₀ and f_e should be small with f₀ to be smaller of the two.

(b) If image A’B’ is exactly at the focus of the eyepiece, then image A”B” is formed at infinity.
If the object AB is very close to the focus of the objective lens of focal length f₀, then magnification M₀ by the objective lens
M_e = \(\frac{L}{f_{0}}\)

where, L is tube length (or distance between lenses L₀ and L_e) Magnification M_e by the eyepiece
M_e = \(\frac{D}{F_{e}} \)
where, D = Least distance of distinct vision
Total magnification, m = M₀ M_e = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 3.
Explain with the help of a labelled ray diagram, how is image formed in an astronomical telescope. Derive an expression for its magnifying power.
Or
Draw a ray diagram showing the image formation of a distant object by a refracting telescope. Define Its magnifying power and write the two important factors considered to increase the magnifying power. Describe briefly the two main limitations and explain how far these can be minimised in a reflecting telescope.
Answer:
Astronomical (Refracti ng) Telescope
Construction: It consists of two co-axial cylindrical tubes, out of which one tube is long and wide, while the other tube is small and narrow. The narrow tube may be moved in and out of the wide tube by rack and pinion arrangement. At one end of wide tube an achromatic convex lens L₁ is placed, which faces the object and is so-called objective (lens). The focal length and aperture of this lens are kept large. The large aperture of objective is taken that it may collect sufficient light to form a bright image of a distant object. The narrow tube is towards eye and carries an achromatic convex lens 12 of small focal length and small aperture on its outer end. This is called eye-lens or eyepiece.

The small aperture of eye lens is taken so that the whole light refracted by it may reach the eye. Cross-wires are fitted at a definite distance from the eye lens. Due to large focal length of objective lens and small focal length of eye lens, the final image subtends a large angle at the eye and hence the object appears large. The distance between the two lenses may be arranged by displacing narrow tube in or out of wide tube by means of rack and pinion arrangement.

Adjustment: First of all the eyepiece is moved backward and forward in the narrow tube and focused on the cross-wires. Then the objective lens is directed towards the object and narrow tube is displaced in or out of wide tube until the image of object is formed on cross-wires and there is no parallax between the image and cross-wires. In this position, a clear image of the object is seen. As the image is formed by refraction of light through both the lenses, this telescope is called the refracting telescope.

Working: Suppose AB is an object whose end A is on the axis of telescope. The objective lens (L₁) forms the image A’B’ of the object AB at its second principal focus F₀.
This image is real, inverted and diminished. This image A’ B’ acts as an object for the eyepiece L₂ and lies between first focus f_e‘ and optical centre C₂ of lens L₂.
Therefore, eyepiece forms its image A” B” which is virtual, erect and magnified.
Thus, the final image A” B” of object AB formed by the telescope is magnified, inverted and lies between objective and eyepiece.

Magnifying Power: The magnifying power of a telescope is measured by the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by object on the eye. i.e.,
Magnifying power M = \(\frac{\beta}{\alpha}\)
As α and β are very small angles, therefore, from figure.

The angle subtended by final image A” B” on eye.
β = angle subtended by image A’B’ on eye
= tanβ = \(\frac{A^{\prime} B^{\prime}}{C_{2} A^{\prime}}\)
As the object is very far (at infinity) from the telescope, the angle subtended by object at eye is same as the angle subtended by object on objective lens.
∴ α = tan α = \(\frac{A^{\prime} B^{\prime}}{C_{1} A^{\prime}}\)
∴ M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} / C_{2} A^{\prime}}{A^{\prime} B^{\prime} / C_{1} A^{\prime}}=\frac{C_{1} A^{\prime}}{C_{2} A^{\prime}}\)
If the focal lengths of objective and eyepiece be f₀ and f_e, distance of image A’B’ from eyepiece be u_e, then by sign convention
C₁A’ = +f₀
C₂A’ = – u_e
∴ M = –\(\frac{f_{o}}{u_{e}}\) ……………………………… (1)

If v_e is the distance of A” B” from eye-piece, then by sign convention, f_e is positive, u_e and v_e are both negative. Hence, by lens formula = \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
we have
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or
\(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Substituting this value in eq. (1), we get
M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\) …………………………. (2)

This is the general formula for magnifying power. In this formula, only numerical values of f₀, f_e and v_e are to be used because signs have already been used.
Length of Telescope : The distance between objective and eyepiece is called the length (L) of the telescope. Obviously,
L = L₁L₂ =C₁C₂ = f₀+u_e …………………… (3)

Now there arise two cases :
(i) When the final image is formed at minimum distance (D) of distinct vision then v_e =D
∴ M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{D}\right)=-\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\) …………………………… (4)
Length of telescope L = f₀ + u_e

(ii) In normal adjustment position, the final image is formed at infinity: For relaxed eye, the final image is formed at infinity. In this state, the image A’B’ formed by objective lens should be at first the principal focus of eyepiece, i.e.,
u_e = f_e and v_e
∴ Magnifying power,
M = – f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{\infty}\right)\) = –\(\frac{f_{o}}{f_{e}}\)
Length of telescope = f₀ + f_e
For large magnifying power, f₀ should be large and f_e should be small. For high resolution of the telescope, diameter of the objective should be large.

Factors for Increasing the Magnifying Power
1. Increasing focal length of objective
2. Decreasing focal length of eyepiece

Limitations
1. Suffers from chromatic aberration
2. Suffers from spherical aberration
3. Small magnifying power
4. Small resolving power

Advantages of Reflecting Telescope
1. No chromatic aberration, because mirror is used.
2. Spherical aberration can be removed by using a parabolic mirror.
3. Image is bright because no loss of energy due to reflection.
4. Large mirror can provide easier mechanical support.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction

Very short answer type questions

Question 1.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the plate oscillate, the changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause. Also, copper being substance, it gets magnetised in the opposite direction, so the plate motion gets damped.

Question 2.
On what factors does the magnitude of the emf induced in the circuit due to magnetic flux depend ?
Answer:
Depends on the time rate of change in magnetic flux (or simply change in Magnetic flux)
\(|\varepsilon|=\frac{\Delta \phi}{\Delta t}\)

Question 3.
A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.
Answer:
When the current begins to grow through the electromagnet, the magnetic flux through the disc begins to increase. This sets up eddy current in the disc in the same direction as that of the electromagnetic current.

Thus, if the upper surface of electromagnetic acquires AT-polarity, the lower surface of the disc also acquires N-polarity. As, same magnetic poles repel each other, the light metallic disc is thrown up.

Question 4.
State the Faraday’s law’ of electromagnetic induction.
Answer:
On the basis of his experiment, Faraday gave the following two laws:
First Law: Whenever magnetic flux linked with a circuit changes, an emf is induced in it which lasts, so long as change in flux continuous.
Second Law: The emf induced in loop or closed circuit is directly proportional to the rate of change of magnetic flux linked with the loop
i.e., ε ∝ \(\frac{(-) d \phi}{d t}\) or ε = -N \(\frac{d \phi}{d t}\)
where, N= number of turns in the coil. Negative sign indicates the Lenz’s law.

Question 5.
State Lenz’s law. A metallic rod held horizontally along East-West direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
Answer:
Lenz’s Law: The direction of the induced emf, or the current, in any circuit is such as to oppose the cause that produces it.

Yes, emf will be induced in the rod as there is change in magnetic flux. When a metallic rod held horizontally along East-West direction, is allowed to fall freely under gravity i.e., fall from North to South, the intensity of magnetic lines of the earth’s magnetic field changes through it, i.e., the magnetic flux changes and hence emf induced in it.

Question 6.
How does the mutual inductance of a pair of coils change, when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?
Answer:
(i) AΦ = MI, with the increase in the distance between the coils the magnetic flux linked with the secondary coil decreases and hence, the mutual inductance of the two coils will decreases with the increase of separation between them.

(ii) Mutual inductance of two coils can be found out by
M = μ₀N₁N₂ Al i.e.,
M ∝ N₁N₂, SO, with the increase in number of turns mutual inductance increases.

Question 7.
Why is the core of a transformer laminated?
Answer:
The core of a transformer is laminated because of preventing eddy current being produced in the core.

Question 8.
How can the self-inductance of a given coil having N number of turns, area of cross-section A and lengths l be increased?
Answer:
The self-inductance can be increased by the help of electric fields. It does not depend on the current through circuit but depends upon the permeability of material from which the core is made up off.

Question 9.
Consider a magnet surrounded by a wire with an on/off switch S (as shown in figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current ?Explain (NCERT Exemplar)

Answer:
No part of the wire is moving and so motional e.m.f. is zero. The magnet is stationary and hence the magnetic field does not change with time. This means no electromotive force is produced and hence no current will flow in the circuit.

Question 10.
A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced emf resist this decrease, which can be done by an increase in current.

Question 11.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced emf should resist this increase, which can be achieved by a decrease in current. However, this change will be momentarily.

Question 12.
Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current /. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring? (NCERT Exemplar)
Answer:
When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase.

Short answer type questions

Question 1.
Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula Φ = B₁ dA₁, B₂ dA₂…. Now, if we choose two different surfaces S₁ and S₂ having C as their edge, would we get the same answer for flux. Justify your answer. (NCERTExemplar)

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let dΦ = BdA represents magnetic lines in an area A to B.

By the concept of continuity of lines B cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂. Therefore, in both the cases we gets the same answer for flux.

Question 2.
What are eddy currents? Write their two applications.
Answer:
Eddy Current: Eddy currents are the currents induced in the bulk pieces of conductors when the amount of magnetic flux linked with the conductor changes.

Eddy currents can be minimised by taking laminated core, consists of thin metallic sheet insulated from each other by varnish instead of a single solid mass. The plane of the sheets should be kept perpendicular to the direction of the currents. The insulation provides high resistance hence, eddy current gets minimised.

Applications
(i) Electromagnetic damping
(ii) Induction furnace.

Question 3.
(i) A rod of length l is moved horizontally with a uniform – velocity v in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.

(ii) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.
Answer:
(i) Consider a straight conductor moving with velocity v and U shaped conductor placed in perpendicular magnetic field as shown in the figure.

Let conductor shifts from ab to a’ b’ in time dt, then change in magnetic flux
dΦ = B × change in area
= B × (areaa’b’ab)
= B × (l × vdt)
∴  \(\frac{d \phi}{d t}\) Bvl
∴  Induced emf lei \(|\varepsilon|=\frac{d \phi}{d t}\) = Bvl

(ii) During motion, free e^– are shifted at one end due to magnetic force so due to polarisation of rod electric field is produced which applies electric force on free e^– on opposite direction.

At equilibrium of Lorentz force,
F_e + F_m = 0
qE + q(v × B) = 0
E = -v × B = B × v
\(|E|=|B v \sin 90|\)
\(\frac{d v}{d r}\) = Bv
P_D = Bvl

Question 4.
(a) How does the mutual inductance of a pair of coils change when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?

(b) A plot of magnetic flux (Φ) versus current (I), is shown in the figure for two inductors A and B. Which of the two has large value of self-inductance?

(c) How is the mutual inductance of a pair of coils affected when
(i) separation between the coils is increased?
(ii) the number of turns in each coil is increased?
(iii) a thin iron sheet is placed between the two coils, other factors remaining the same?
Justify your answer in each case.

Answer:
(a)
(i) Mutual inductance decreases.
(ii) Mutual inductance increases.
Concept
(i) If distance between two coils is increased as shown in figure.

It causes decrease in magnetic flux linked with the coil C₂. Hence induced emf in coil C₂ decreases by relation ε₂ = \(\frac{-d \phi_{2}}{d t}\). Hence mutual inductance decreases.
(ii) From relation M₂₁ = μ₀N₁N₂ Al, if number of turns in one of the coils or both increases, means mutual inductance will increase.

(b) Φ = LI ⇒ \(\frac{\phi}{I}\) = L
The slope of \(\frac{\phi}{I}\) of straight line is equal to self-inductance L. It is larger for inductor A; therefore inductor A has larger value of self inductanc ‘ L’.

(c)
(i) When the relative distance between the coil is increased, the leakage
of flux increases which reduces the magnetic coupling of the coils. So magnetic flux linked with all the turns decreases. Therefore, mutual inductance will be decreased.

(ii) Mutual inductance for a pair of coil is given by
M = K\(\sqrt{L_{1} L_{2}}\)
where, L = \(\frac{\mu N^{2} A}{l}\) and L is called self inductance. Therefore, when the number of turns in each coil increases, the mutual inductance also increases.

(iii) When a thin iron sheet is placed between the two coils, the mutual inductance increases because M ∝ permeability. The permeability of the medium between coils increases.

Question 5.
Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, It takes more time to come down than It takes for a similar
unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain. (NCERT Exemplar)
Answer:
For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the motion of the magnet. Therefore magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy
currents and will fall an acceleration g. Thus the magnet will take more time.

Question 6.
A magnetic field B = B₀ sin(ωt) k̂ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity y, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity? (NCERT Exemplar)

Answer:
Let us assume that the parallel wires are at y = 0 i. e., along x-axis and y = d. At t = 0, AB has x = 0, i. e., along y-axis and moves with a velocity v. Let at time t, wire is at x (t) = vt.
Now, the motional emf across AB is
= (B₀sinωt) vd(-ĵ)
emf due to change in field (along OBAC)
= -B₀ωcosωt (t)d
Total emf in the circuit = emf due to change in field (along OBAC) + the motional emf across AB = -B₀d[ωxcos(ωt) + vsin (ωt)]
Electric current in clockwise direction is given by,
= \(\frac{B_{0} d}{R}\) = (ωxcosωt + vsinωt)
The force acting on the conductor is given by F = ilB sin 90° = ilB
Substituting the values, we have
Force needed along i = \(\frac{B_{0} d}{R}\) (ωx cos ωt + vsinωt) × d × B sinωt
= \(\frac{B_{0}^{2} d^{2}}{R}\)(ωx cos ωt + vsinωt) sinωt
This is the required expression for force.

Long answer type questions

Question 1.
(i) How is magnetic flux linked with the armature coil changed in a generator ?
(ii) Derive the expression for maximum value of the induced emf and state the rule that gives the direction of the induced emf.
(iii) Show the variation of the emf generated versus time as the armature is rotated with respect to the direction of the magnetic fields.
Answer:
(i) The direction of flow of current in resistance R get changed alternatively after every half cycle.
Thus, AC is produced in coil.

(ii) Let at any instant total magnetic flux linked with the armature coil is G. and θ = ωt is the angle made by area vector of coil with magnetic field.
Φ = NBA cosθ = NBA cosωt
\(\frac{d \phi}{d t}\) = -NBAω sin ωt
– \(\frac{d \phi}{d t}\) = NBAω sin ωt
By Faraday’s law of emf, e = \(\frac{-d \phi}{d t}\)
Induced emf in coil is given by,
e = NBAω sinωt
e = e₀ sinωt
where, e₀ = NBAω = peak value of induced emf

(iii) The mechanical energy spent in rotating the coil in magnetic field appears in the form of electrical energy.

Question 2.
State the working of AC generator with the help of a labelled diagram.
The coil of an AC generator having N turns, each of area A, is rotated with a constant angular velocity to. Deduce the expression for the alternating emf generated in the coil.
What is the source of energy generation in this device?
Answer:
AC Generator: A dynamo or generator is a device which converts mechanical energy into electrical energy.

Principle: It works on the principle of electromagnetic induction. When a coil rotates continuously in a magnetic field, the effective area of the coil linked normally with the magnetic field lines, changes continuously with time. This variation of magnetic flux with time results in the production of an alternating emf in the coil.

Construction: It consists of the four main parts
(i) Field magnet: It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet.

(ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. It can revolve an axle between the two poles of the field magnet. The drum or ring serves the two purposes: (a) It serves as a support to coils and (b) It increases the magnetic field due to air core being replaced by an iron core.

(iii) Slip rings: The slip rings R₁ and R₂ are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature.

(iv) Brushes: There are two flexible metal plates or carbon rods (B₁ and B₂) which are fixed and constantly touch the revolving rings. The output current in external load RL is taken through these brushes.

Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B₁R_lB₂. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B₂R_LB₁ Thus the direction of induced emf and current changes in the external circuit after each half revolution.

Expression for Induced emf: If N is number of turns in coil, f the frequency of rotation, A area of coil and B the magnetic induction, then induced emf
e = – \(\frac{d \phi}{d t}\) = –\(\frac{d}{d t}\) {NBA (cos 2π ft)} dt dt
= 2π NBA f sin 2π ft
Obviously, the emf produced is alternating and hence the current is also alternating.
Current produced by an AC generator cannot be measured by moving coil ammeter; because the average value of AC over full cycle is zero.
The source of energy generation is the mechanical energy of rotation of armature coil.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics

Very short answer type questions

Question 1.
Define the term ‘wavefront’.
Answer:
It is defined as the locus of all points in a medium vibrating in the same phase.

Question 2.
State Huygen’s principle of diffraction of light.
Answer:
When a wavefront strikes to the corner of an obstacle, lightwave bends around the corner because every point on the wavefront again behaves like a . light source and emit secondary wavelets in all directions (Huygen’s wave theory) including the region of geometrical shadow. This explains diffraction.

Question 3.
Define the term ‘coherent sources’ which are required to produce interference patterns in Young’s double-slit experiment.
Answer:
Two monochromatic sources, which produce light waves, having a constant phase difference are known as coherent sources.

Question 4.
Define Doppler’s effect in light.
Answer:
It states that whenever there is a relative motion between the observer and the source of light, the apparent frequency of light received by the observer is different from the actual frequency of the light emitted by the source of light.

Question 5.
Define Doppler shift.
Answer:
It is defined as the apparent change in the frequency or wavelength of light due to the relative motion between the source and the observer.

Question 6.
Define redshift.
Answer:
It is defined as the shifting of radiations from the source of light towards the red end of the spectrum when the source moves away from the stationary observer. The wavelength increases due to redshift.

Question 7.
Define limit of resolution of an optical instrument.
Answer:
It is defined as the minimum distance by which the timepoint objects are separated so that their images can be seen as just separated by the optical instrument.

Question 8.
Define resolving power of the optical instruments.
Answer:
It is defined as the reciprocal of the limit of resolution of the optical instrument.

Question 9.
How are resolving power of a telescope change by increasing or decreasing the aperture of the objective?
Answer:
We know that the resolving power of telescope is given by
R.P. = \(\frac{D}{1.22 \lambda}\)
As R.p. ∝ D, so by increasing or decreasing D (aperture) of the objective, the resolving power is increased or decreased.

Question 10.
Which of the following waves can be polarised (i) Heat waves (ii) Sound waves? Give reason to support your answer.
Answer:
Heatwaves are transverse or electromagnetic in nature whereas sound waves are not. Polarisation is possible only for transverse waves.

Question 11.
How is linearly polarised light obtain by the process of scattering of light? find the Brewster angle for air-glass interface, when the refractive index of glass = 1.5
Answer:
According to Brewster law
tan i_B = μ
i_B = tan^-1 (μ)
i_B = tan^-1(l. 5)
i_B = 56.30

Question 12.
A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain. (NCERT Exemplar)
Answer:
Only in the special cases when the pass axis of (III) is parallel to (I) or (II), there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (I) is no longer perpendicular to the pass axis of (III).

Question 13.
What is the shape of the wavefront of earth for sunlight? (NCERT Exemplar)
Answer:
Spherical with huge radius as compared to the earth’s radius so that it is almost a plane.

Question 14.
Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
The focal point of a convergent lens is the position of real image formed by this lens when object is at infinity. When another convergent lens of short focal length is placed on the other side, the combination will form a real point image at the combined focus of the two lenses. The wavefronts emerging from the final image will be spherical.

Short answer type questions

Question 1.
State two conditions required for obtaining coherent sources. In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally spaced.
Answer:
Conditions for obtaining coherent sources:
(i) Coherent sources of light should be obtained from a single source by same device.
(ii) The two sources should give monochromatic light.
The separation between the centres or two consecutive bright fringes is the width of a dark fringe.

Hence, all bright and dark fringes are equally spaced on screen.

Question 2.
How will the interference pattern in Young’s double-slit experiment get affected, when
(i) distance between the slits S₁ and S₂ reduced and
(ii) the entire set-up is immersed in water? Justify your answer in each case.
Answer:
(i) The fringe width of interference pattern increases with the decrease in separation between S₁S₂ as
β ∝ \(\frac{1}{d}\)
(ii) The fringe width decrease as wavelength gets reduced when interference set up is taken from air to water.

Question 3.
What is the minimum angular separation between two stars, if a telescope is used to observe them with an objective of aperture 0.2 m? The wavelength of light used is 5900 A.
Answer:
Here, D = diameter of the objective of telescope = 0.2 m
λ = Wavelength of light used = 5900 Å = 5900 x 10^-10 m
Let dθ = Minimum angular separation between two stars =?
Using the relation,
dθ = \(\frac{1.22 \lambda}{D}\) , we get
dθ = \(\frac{1.22 \times 5900 \times 10^{-10}}{0.2}= \) = 3.6 x 10^-6 rad.

Question 4.
Distinguish between polarised and unpolarised light. Does the intensity of polarised light emitted by a polaroid depend on its orientation? Explain briefly. The vibrations in a beam of polarised light make an angle of 60° with the axis of the polaroid sheet. What percentage of light is transmitted through the sheet?
Answer:
A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light. The light from the sun, an incandescent bulb or a candle is unpolarised. If the electric field vector of a light wave vibrates just in one direction perpendicular to the direction of wave propagation, then it is said to be polarised or linearly polarised light.

Yes, the intensity of polarised light emitted by a polaroid depends on orientation of polaroid. When polarised light is incident on a polaroid, the resultant intensity of transmitted light varies directly as the square of the cosine of the angle between polarisation direction of light and the axis of the polaroid.

I ∝ cos² θ or I = I₀ cos² θ
where I₀ = maximum intensity of transmitted light;
θ = angle between vibrations in light and axis of polaroid sheet.
or I =I₀ cos² 60° = \(\frac{I_{0}}{4}\)
Percentage of light transmitted = \(\frac{I}{I_{0}} \) x 100 = \(\frac{1}{4}\) x 100 = 25%

Question 5.
Find an expression for intensity of transmitted light, when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?
Answer:
Let us consider two crossed polarizers, P₁ and P₂ with a polaroid sheet P₃ placed between them.

Let I₀ be the intensity of polarised light after passing through the first polarizer P₁.
If θ is the angle between the axes of P₁ and P₃, then the intensity of the polarised light after passing through P₃ will be I =I₀ cos²θ.
As P₁ and P₂ are crossed, the angle between the axes of P₁ and P₂ is 90°.
∴ The angle between the axes of P₂ and P₃ is (90° – 0).
The intensity of light emerging from P₂ will be given by

The intensity of polarised light transmitted from P₂ will be maximum, when ,
sin 2θ = maximum = 1
⇒ sin2θ = sin9O°
⇒ 2θ = 90°
⇒ θ = 45°
Also, the maximum transmitted intensity will be given by I = \(\frac{I_{0}}{4}\)

Question 6.
State Brewster’s law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.
Answer:
Brewster’s Law: When unpolarized light is incident on the surface separating two media at polarising angle, the reflected light gets completely polarised only when the reflected light and the refracted light are perpendicular to each other. Now, refractive index of denser (second) medium with respect to rarer (first) medium is given by μ = tan i_B, where i_B = polarising angle.
Since refractive index is different for different colours (wavelengths), Brewster’s angle is different for different colours.

Question 7.
Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index? (NCERT Exemplar)
Answer:
When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle
i.e., tani_B = ¹μ₂ = \(\frac{\mu_{2}}{\mu_{1}}\) where μ² < μ¹
when the light rays travels in such a medium, the critical angle is
sin i_c = \(\frac{\mu_{2}}{\mu_{1}}\)
where, μ₂ < μ₁
As | tani_B| > | sin i_C| for large angles i_B <i_C.
Thus, the polarisation by reflection occurs definitely.

Question 8.
Consider a two-slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O. (NCERT Exemplar)

Answer:
From the given figure of two-slit interference arrangements, we can write

The minima will occur when S₂P – S₁P = (2 n -1)\(\frac{\lambda}{2}\)
i.e., [D² +(D + X)²]^1/2 -[D² + (D -x)²]^1/2
= \(\frac{\lambda}{2}\)
[for first minima n = 1]
If x = D
We can write [D² +4D²]^1/2 -[D² +0]^1/2 = \(\frac{\lambda}{2}\)
⇒ [5D²]^1/2 – [D²]^1/2 = \(\frac{\lambda}{2}\)
⇒ \(\sqrt{5}\)D – D = \(\frac{\lambda}{2}\)

Long answer type questions

Question 1.
In Young’s double-slit experiment, deduce the conditions for (i) constructive, and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position ‘X’ on the screen.
Answer:
Conditions for Constructive and Destructive Interference :
When two waves of same frequency and constant initial phase difference travel in the same direction along a straight line simultaneously, they superpose in such a way that the intensity of the resultant wave is maximum at certain points and minimum at certain other points. The phenomenon of redistribution of intensity due to superposition of two waves of same frequency and constant initial phase difference is called the interference.

The waves of same frequency and constant initial phase difference are called coherent waves. At points of medium where the waves arrive in the same phase, the resultant intensity is maximum and the interference at these points is said to be constructive. On the other hand, at points of medium where the waves arrive in opposite phase, the resultant intensity is minimum and the interference at these points is said to be destructive. The positions of maximum intensity are called maxima while those of minimum intensity are called minima. The interference takes place in sound and light both.

Mathematical Analysis: Suppose two coherent waves travel in the same direction along a straight line, the frequency of each wave is \(\frac{\omega}{2 \pi}\) and amplitudes of electric field are a₁ and a₂ respectively. If at any time t, the electric fields of waves at a point are y₁ and y₂ respectively and phase difference is, Φ then equation of waves may be expressed as
y₁ = a₁ sin ωt ………………………. (1)
y₂ = a₂ sin ωt +Φ) ……………………………………….. (2)
According to Young’s principle of superposition, the resultant displacement at that point will be
y = y₁+y₂ ……………………………….. (3)
Substituting values of y₁ and y₂ from (1) and (2) in (3), we get
y = a₁ sin ωt + a₂ sin(ωt + Φ)

Using trigonometric relation,
sin(ωt +Φ) = sinωtcosΦ +cosωtsinΦ
y = a₁ sin ωt + a₂(sinωtcosΦ) + cosωt sin Φ)
= (a₁ +a₂cos Φ) sinωt + (a₂ sinΦ)cosωt …………………………….. (4)
Let a₁ + a₂ cosΦ = A cos θ ……………………………………… (5)
and a₂ sinΦ = A sinθ ………………………………………… (6)

Where A and θ are new constants
Then equation (4) gives
y = A cosθ sinωt + A sinθ cosωt
= A sin (ωt +θ) ……………………………………………. (7)
This is the equation of the resultant disturbance. Clearly the amplitude of resultants disturbance is A and phase difference from first wave is 0. The values of A and 0 are determined by (5) and (6). Squaring (5) and (6) and then adding, we get

∴ Amplitude,
A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi} \) …………………………… (8)
As the intensity of a wave is proportional to its amplitude in arbitrary units I = A²
∴ Intensity of resultant wave,
I = A² = a₁² + a₂² + 2a₁a₂ cosΦ ……………………….. (9)
Clearly, the intensity of the resultant wave at any point depends on the amplitudes of individual waves and the phase difference between the waves at the point.

Constructive Interference: For maximum intensity at any point
cos Φ = +1
or phase difference Φ = 0,2π,4π,6π,……………………….
= 2nπ (n=0,1,2,3,……………………) …………………………………… (10)
The maximum intensity
I_max = a¹₂+a_2²
= (a₁+a₂)² …………………………..(11)
Path difference
Δ = \(\frac{\lambda}{2 \pi}\) x phase difference
= \(\frac{\lambda}{2 \pi} \) x 2nπ …………………………………………. (12)
Clearly, the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of π or path difference between them is the integral multiple of λ and maximum intensity is (a₁ +a₂)²

which is greater than the sum intensities of individual waves by an amount 2a₁a₂.
Destructive Interference : For minimum intensity at any point CosΦ = -1
or phase difference,
Φ = π,3π,5π,7π, …………………………..
– (2n-l)π, n = 1,2,3,… …………………………………. (13)
In this case the minimum intensity,
I_min =a₁² +a₂² – 2a₁a₂
= (a₁-a₂)² ………………………… (14)

Path difference, Δ = \(\frac{\lambda}{2 \pi}\) x Phase difference
= \(\frac{\lambda}{2 \pi}\) x (2n – 1)π
= (2n-l) \(\frac{\lambda}{2}\)

Clearly, the minimum intensity is obtained in the region of superposition at those points where waves meet in opposite phase or the phase difference between the waves is odd multiple of π or path difference between the waves is odd multiple of \(\frac{\lambda}{2}\) and minimum intensity = (a₁ -a₂)² which is less than the sum of intensities of the individual waves by an amount 2a₁a₂.

From equations (12) and (14) it is clear that the intensity 2a₁a₂ is transferred from positions of minima to maxima, this implies that the interference is based on conservation of energy i.e., there is no wastage of energy.
Variation of Intensity of light with position x is shown in fig.

Question 2.
Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle θ in single slit diffraction.
Answer:
Diffraction of Light at a Single Slit: When monochromatic light is made incident on a single slit, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains bright and dark bands, the intensity of central band is maximum and goes on decreasing on both sides.

Explanation: Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. According to Fresnel, the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.

Let θ be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. The path difference between rays diffracted at points A and B,
Δ = BP – AP = BN
In ΔANB, ∠ANB = 90°
and ∠BAN = θ
∴ sinθ = \(\frac{B N}{A B}\) or BN = AB sinθ
As AB = width of slit = a
Path difference Δ = asinθ ……………………………… (1)

To find the effect of all coherent waves at P, we have to sum up their contribution, each with a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below :
At the central point C of the screen, the angle 0 is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C. If point P on screen is such that the path difference between rays starting from edges A and B is λ, then path difference,

If angle θ is small,
sinθ = θ = \(\frac{\lambda}{a}\) ……………………………. (2)
Minima: Now we divide the slit into two equal halves AO and OB, each of width \(\frac{a}{2}\).
Now for every point, M₁ in AO, there is a corresponding point M₂ in OB, such that M₁M₂ = \(\frac{a}{2}\) .
Then path difference between waves arriving at P and starting from M₁ and M₂ will be \(\frac{a}{2}\) sin θ = \(\frac{\lambda}{2}\).

This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. Thus equation (2) gives the angle of diffraction at which intensity falls to zero. Similarly it may be shown that the intensity is zero for sin θ = \(\frac{n \lambda}{a}\) , with n as integer. Thus, the general condition of minima is asinθ = nλ ……………………………………… (3)

Secondary Maxima: Let us now consider angle θ such that
sin θ = θ = \(\frac{3 \lambda}{2 a}\)

Which is midway between two dark bands given by
sin θ = θ = \(\frac{\lambda}{a} \) and sin θ = θ = \(\frac{2 \lambda}{a}\)
Let us now divide the slit into three parts. If we take the first two parts of slit, the path difference between rays diffracted from the extreme ends of the first two parts.
\(\frac{2}{3}\) a sin θ = \(\frac{2}{3} a \times \frac{3 \lambda}{2 a}\) = λ

Then the first two parts will have a path difference of \(\frac{\lambda}{2}\) and cancel the effect of each other. The remaining third part will contribute to the intensity at a point between two minima. Clearly, there will be maxima between first two minima, but this maximum will be of much weaker intensity than central maximum.

This is called first secondary maxima. In a similar manner, we can show that there are secondary maxima between any two consecutive minima; and the intensity of maxima will go on decreasing with increase of order of maxima.
In general, the position of nth maxima will be given by
a sin θ = \(\left(n+\frac{1}{2}\right)\) λ (n =1, 2, 3, 4,…) ………………………………… (4)
The intensity of secondary maxima decreases with increase of order n because with increasing n, the contribution of slit decreases.
For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter

Very short answer type questions

Question 1.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
The angle of dip is given by
θ = tan^-1 (\(\frac{B_{V}}{B_{H}}\))
B_V = vertical component of the earth’s magnetic field.
B_H = horizontal component of the earth’s magnetic field.
So, as B_V = B_H
Then, θ = tan^-1 (1) = 45°
∴ The angle of dip will be θ = 45°.

Question 2.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the copper plates oscillate in the magnetic field between the two plates of the magnet, there is a continuous change of magnetic flux linked with the pendulum. Due to this, eddy currents are set up in the copper plate which try to oppose the motion of the pendulum according to the Lenz’s law and finally bring it to rest.

Question 3.
Relative permeability of a material μ_r = 0.5. Identify the nature of the magnetic material and write its relation of magnetic susceptibility.
Answer:
The nature of magnetic material is a diamagnetic.
μ_r = 1 + χ_m

Question 4.
Which of the following substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni
Answer:
Diamagnetic substances are (i) Bi (ii) Cu.

Question 5.
The susceptibility of a magnetic material is -4.2 × 10^-6. Name the type of magnetic material, it represents.
Answer:
Negative susceptibility represents diamagnetic substance.

Question 6.
What are permanent magnets? Give one example.
Answer:
Substances that retain their attractive property for a long period of time at room temperature are called permanent magnets.
Examples: Those pieces which are made up of steel, alnico, cobalt and ticonal.

Question 7.
Why is the core of an electromagnet made of ferromagnetic materials?
Answer:
Ferromagnetic material has a high retentivity. So on passing current through windings it gains sufficient magnetism immediately.

Question 8.
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism. (NCERT Exemplar)
Answer:
Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this alignment is disturbed and hence susceptibilities of both decrease as temperature increases.

Question 9.
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet.
(i) In which direction will it move?
(ii) What will be the direction of its magnetic moment? (NCERT Exemplar)
Answer:
When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.
(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

Question 10.
Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q. (NCERT Exemplar)
Answer:
In adjoining figure:

(i) P is in S (needle will point both north)
Declination = 0
P is also on magnetic equator.
∴ Dip = 0

(ii) Q is on magnetic equator.
∴ Dip = 0
But declination = 11.3.

Short answer type questions

Question 1.
Explain the following:
(i) Why do magnetic field lines form continuous closed loops?
(ii) Why are the field lines repelled (expelled) when a
diamagnetic material is placed in an external uniform magnetic field?
Answer:

(i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result, the net magnetic flux of a magnet is always zero.
(ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is induced in opposite direction. So, magnetic lines of force are repelled.

Question 2.
How does a circular loop carrying current behaves as a magnet?
Answer:
The current round in the face of the coil is in anti-clockwise direction, then this behaves like a North pole, whereas when it viewed from other scale, then current round in it is in clockwise direction necessarily forming South pole of magnet.

Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole.

Question 3.
Give two points to distinguish between a paramagnetic and diamagnetic substance.
Answer:

Paramagnetic substance	Diamagnetic substance
1. A paramagnetic substance is feebly attracted by magnet.	A diamagnetic substance is feebly repelled by a magnet.
2. For a paramagnetic substance, the intensity of magnetisation has a small positive value.	For a diamagnetic substance, the intensity of magnetism has a small negative value.

Question 4.
(a) How is an electromagnet different from a permanent magnet?
Write two properties of a material which makes it suitable for making (i) a permanent magnet, and (ii) an electromagnet.
Answer:
(a) An electromagnet consists of a core made of a ferromagnetic material placed inside a solenoid. It behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.

A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetised one.

(b) Properties of material are as below:
(i) Permanent magnet

• Retentivity and coercivity should be large
• Magnetically hard

(ii) An electromagnet

• Magnetically soft
• Coercivity should be low.

Question 5.
A bar magnet of magnetic moment M and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscifiations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T’ for each piece? (NCERT Exemplar)
Answer:
Given, I = moment of inertia of the bar magnet
m = mass of bar magnet
l = length of magnet about an any passing through its centre and perpendicular to its length
M = magnetic moment of the magnet
B = uniform magnetic field in which magnet is oscillating, we get time period of oscillation is
T = 2π\(\sqrt{\frac{I}{M B}}\)
Here I = \(\frac{m l^{2}}{12}\)
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is

Question 6.
A uniform conducting wire of length 12 a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V₀. Find the magnetic moment of the coils in each case. (NCERT Exemplar)

(i) Area of equilateral triangle, A = \(\frac{\sqrt{3}}{4}\) a²
(ii) Area of square, A = a²

Long answer type questions

Question 1.
Verify the Ampere’s law for magnetic field of a point dipole of dipole moment M = Mk̂. Take C as the closed curve running clockwise along
(i) the z-axis from z a > 0 to z = R,
(ii) along the quarter circle of radius R and centre at the origin in the first quadrant of vz-plane,
(iii) along the x-axis from x = R to x = a, and
(iv) along the quarter circle of radius a and centre at the origin
in the first quadrant of xz-plane (NCERT Exemplar)
Answer:
From P to Q, every point on the z-axis lies at the axial line of magnetic
dipole of moment \(\vec{M}\). Magnetic field induction at a point distance z from the magnetic dipole of moment is

(ii) Along the quarter circle QS of radius R as-.given in the figure below

The point A lies on the equatorial line of the magnetic dipole of moment M sin0. Magnetic field at point A on the circular arc is
B = \(\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{R^{3}}\) ; \(\overrightarrow{d l}\) = Rdθ

(iii) Along x-axis over the path ST, consider the figure given below.

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance x from the dipole is

(iv) Along the quarter circle TP of radius a. Consider the figure given below

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter

Very short answer type questions

Question 1.
In the photoelectric effect, why should the photoelectric current increase as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain.
Answer:
The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the photosensitive surface.

Question 2.
Define the term ‘threshold frequency’ in relation to photoelectric effect.
Answer:
Threshold frequency is defined as the minimum frequency of incident radiation which can cause photoelectric emission. It is different for different metal.

Question 3.
WrIte the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.
Answer:
Features of the photons are as:

• Photons are particles of light having energy E = hv and momentum p = \(\frac{h}{\lambda}\), where h is Planck’s constant.
• Photons travel with the speed of light in vacuum, independent of the frame of reference.
• Intensity of light depends on the number of photons crossing unit area in a unit time.

Question 4.
Define Intensity of radiation on the basis of photon picture of light. Write its SI unit.
Answer:
The amount of light energy or photon energy incident per meter square per second is called intensity of radiation1 Its SI unit is \(\frac{\mathrm{W}}{\mathrm{m}^{2}}\) or J/s m

Question 5.
State de Broglie hypothesis.
Answer:
According to the hypothesis of de Brogue “The atomic particles of matter moving with a given velocity, can display the wave-like properties.” i.e., λ = \(\frac{h}{m v}\) (mathematically)

Question 6.
Write the relationship of de Brogue wavelength λ associated with a particle of mass m terms of its kinetic energy E.
Answer:
Kinetic energy E_K = \(\frac{p^{2}}{2 m}\)
where, p = momentum
m = mass and E_K = kinetic energy
⇒ p = \(\sqrt{2 m E_{K}} \)
de Brogue wavelength,
λ = \(\frac{h}{p}\)
Where, p = \(\sqrt{2 m E_{K}} \)
⇒ λ = \(\frac{h}{\sqrt{2 m E_{K}}}\)

Question 7.
Name the phenomenon which shows the quantum nature of electromagnetic radiation.
Answer:
Photoelectric effect.

Question 8.
Do all the electrons that absorb a photon comes out as photoelectrons? (NCERT Exemplar)
Answer:
No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

Question 9.
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelengths and emit light of shorter wavelengths? (NCERT Exemplar)
Answer:
In the first case, energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances.

Question 10.
There are two sources of light, each emitting with a power 100W.
One emits X-rays of wavelength 1 nm and the other visible light at 500 nm.
Find the ratio of number of photons of X-rays the photons of visible light of the given wavelength. (NCERT Exemplar) Ans. Total E is constant.
Let n₁ and n₂ be the number of photons of X-rays and visible region.
n₁E₂ = n₂E₂

Short answer type questions

Question 1.
What is meant by work function of a metal? How does the value of work function influence the kinetic energy of electrons liberated during photoelectron emission?
Answer:
Work Function: The minimum energy required to free an electron from metallic surface is called the work function.
Smaller the work function, larger the kinetic energy of emitted electron.

Question 2.
Show mathematically how Bohr’s postulate of quantization of orbital angular momentum in hydrogen atom is explained by de Broglie’s hypothesis.
Answer:
According to de Broglie’s hypothesis,
λ = \(\frac{h}{m v}\) …………………………… (1)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelength.
2πr = nλ …………………………………… (2)

Substituting value of λ from (1) in (2), we get
2πr = n\(\frac{h}{m v}\)
⇒ mvr = n\(\frac{h}{2 \pi}\) ……………………………………… (3)
This is Bohr’s postulate of quantization of energy levels.

Question 3.
Write briefly the underlying principle used in the Davison-Germer experiment to verify wave nature of electrons experimentally. What is the de Broglie wavelength of an electron with kinetic energy (E_K) 120 eV?
Answer:
Principle: Diffraction effects are observed for beams of electrons scattered by the crystals.
λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E_{K}}}=\frac{h}{\sqrt{2 m e V}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 120}}\)
λ = 0.112 nm.

Question 4.
(i) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(ii) Discuss briefly how wave theory of light cannot explain these features.
Answer:
(i) Three experimentally observed features in the phenomenon of photoelectric effect are as follows :

• Intensity: When intensity of incident light increases as one photon ejects one electron, the increase in intensity will increase the number of ejected electrons. Frequency has no effect on photoelectron.
• Frequency: When the frequency of incident photon increases, the kinetic energy of the emitted electrons increases. Intensity has no effect on kinetic energy of photoelectrons.
• No Time Lag: When energy incident photon is greater than the work function, the photoelectron is immediately ejected. Thus, there is no time lag between the incidence of light and emission of photoelectrons.

(ii) These features cannot be explained in the wave theory of light because wave nature of radiation cannot explain the following :

• The instantaneous ejection of the photoelectrons.
• The existence of threshold frequency for a metal surface.
• The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

Question 5.
A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions :
(i) Do the emitted photoelectrons have the same kinetic energy?
(ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?
(iii) On what factors does the number of emitted photoelectrons depend?
Answer:
In photoelectric effect, an electron absorbs a quantum of energy hv of radiation, which exceeds the work function, an electron is emitted with maximum kinetic energy.
E_{K max} = hv – W
(i) No, all electrons are bound with different forces in different layers of the metal. So, more tightly bound electron will emerge with less kinetic energy. Hence, all electrons do not have same kinetic energy.
(ii) No, because an electron cannot emit out if quantum energy hv is less than the work function of the metal. The K_E depends on the energy of each photon.
(iii) Number of emitted photoelectrons depends on the intensity of the radiations provided the quantum energy hv is greater than the work function of the metal.

Question 6.
Define the term “cut-off frequency” in photoelectric emission. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photoelectrons is v₁. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photoelectrons is v₂. Find the ratio v₁: v₂.
Answer:
Cut-off Frequency: It is that maximum frequency of incident radiation below which no photoemission takes place from a photoelectric material. According to Einstein’s photoelectric equation
E_{K max} = \(\frac{h c}{\lambda}-\phi \) = \(h v-\phi\)
Given that threshold frequency of the metal is f. If light of frequency, 2f is incident on metal plate, maximum velocity of photoelectron is v₁ then,
\(\frac{1}{2} m v_{1}^{2}\) = h (2f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = hf …………………….. (1)

If light of frequency, 5f is incident and maximum velocity of photoelectron is v₂.
\(\frac{1}{2} m v_{1}^{2}\) = h(5f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = 4hf ………………………………… (2)
Dividing (1) by (2), we get
\(\left(\frac{v_{1}}{v_{2}}\right)^{2}=\frac{1}{4}\)
⇒ \(\frac{v_{1}}{v_{2}}=\frac{1}{2}\)
∴ v₁ = v₂ = 1:2

Question 7.
Two monochromatic beams A and B of equal intensity I hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies? (NCERTExemplarl
Answer:
Let no. of photons falling per second of beam A = n_A
No. of photons falling per second of beam B = n_B
Energy of beam A = hv_A
Energy of beam B = h v_B

According to question, I = n_Av_A = n_Bv_B
\(\frac{n_{A}}{n_{B}}=\frac{v_{B}}{v_{A}}\) or \(\frac{2 n_{B}}{n_{B}}=\frac{v_{B}}{v_{A}}\)
⇒ v_B = 2 v_A
The frequency of beam B is twice that of A.

Question 8.
Consider Fig. for photoemission.

How would you reconcile with momentum- conservation? No light (Photons) have momentum in a different direction than the emitted electrons. (NCERT Exemplar)
Answer:
The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons.

Long answer type questions

Question 1.
Describe Davisson and Germer’s experiment to demonstrate the wave nature of electrons. Draw a labeled diagram of apparatus used.
Answer:
Davisson and Germer Experiment: In 1927 Davisson and Germer performed a diffraction experiment with electron beam in analogy with X-ray diffraction to observe the wave nature of matter.
Apparatus: It consists of three parts
(i) Electron gun : It gives a fine beam of electrons, de Brogue used electron beam of energy 54 eV. de Brogue wavelength associated with this beam

λ = \(\frac{h}{\sqrt{2 m E_{k}}}\)
Here, m = mass of electron = 9.1 x 10^-31 kg
E_K = Kinetic energy of electron = 54eV
= 54 x 1.6 x 10^-19 J = 86.4 x 10^-19 J
∴ λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 86.4 \times 10^{-19}}}\)
= 1.66 x 10^-10 = 1.66 Å
(ii) Nickel crystal: The electron beam was directed on nickel crystal against the electron detector. The smallest separation between nickel atoms is O.914Å. Nickel crystal behaves as diffraction grating.

(iii) Electron detector: It measures the intensity of electron beam diffracted from nickel crystal. It may be an ionization chamber fitted with a sensitive galvanometer. The energy of electron beam, the angle of incidence of beam on nickel crystal and the position of detector can all be varied.

Method: The crystal is rotated in small steps to change the angle (α say) between incidence and scattered directions and the corresponding intensity (I) of scattered beam is measured. The variation of the intensity (I) of the scattered electrons with the angle of scattering a is obtained for different accelerating voltages.

The experiment was performed by varying the accelerating voltage from 44 V to 68 V. k was noticed that a strong peak appeared in the intensity (I) of the scattered electron for an accelerating voltage of 54 V at a scattering angle α = 50°.

From Bragg’s law, 2d sinθ = nλ
Here, n =1,d =0.914 Å,θ =65°
∴ λ = \(\frac{2 d \sin \theta}{n}=\frac{2 \times(0.914 \AA) \sin 65^{\circ}}{1}\)
=2 x 0.914 x 0.9063Å =1.65Å
The measured wavelength is in close agreement with the estimated de Broglie wavelength. Thus the wave nature of electrons is verified. Later on G.P. Thomson demonstrated the wave nature of fast electrons. Due to their work Davisson and G.P. Thomson were awarded Nobel Prize in 1937.