PSEB 11th Class Chemistry Solutions Chapter 2 Structure of Atom

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 2 Structure of Atom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

PSEB 11th Class Chemistry Guide Structure of Atom InText Questions and Answers

Question 1.
(i) Calculate the number of electrons that will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Mass of one electron = 9.1 × 10^-31 g
Number of electrons in 1.0 g = \(\frac{1}{9.1 \times 10^{-31}}\) = 1.098 x 10³⁰ electrons

(ii) One mole of electron = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 1023 electrons – (9.1 x 10^-31 kg) × (6.022 x 10²³)
= 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole of electrons = 1.602 × 10^-19 × 6.022 × 10²³
= 9.65 × 10⁴ coulomb.

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C.
(Assume that mass of a neutron = 1.675 × 10^-27 kg)
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed?
Answer:
(i) Number of electrons present in 1 molecule of methane (CH₄)
= {1(6) + 4(1)} = 10 electrons
Number of electrons present in 1 mole i.e., 6.022 × 10²³ molecules of
methane = 6.022 × 10²³ × 10 = 6.022 × 10²⁴ electrons
(ii) (a) Mass of a neutron = 1.675 × 10^-27 kg
14.0 g of ¹⁴C contains 1 mole of atoms of C-14
∴ 14.0 g of 14C contains 8 × 6.022 × 10²³ neutrons
[∵ each atom of C-14 contains 8 neutrons]
7 mg (= 7 × 10^-3 g) of ¹⁴C contains
= \(\frac{6.022 \times 10^{23} \times 8 \times 7 \times 10^{-3}}{14}\) = 2.4092 × 10²¹ neutrons

(b) Mass of one neutron = 1.675 × 10²⁷ kg
Mass of 2.4092 × 10²¹ neutrons
= (2.4092 × 10²¹)(1.675 × 10^-27 kg) = 4.035 × 10^-6 kg

(iii) 1 mole of NH₃ contains protons = 7 + 3 = 10 moles of protons
= 6.022 × 10²³ × 10 protons = 6.022 × 10²⁴ protons

(a) 17 gNH₃ contains 6.022 × 10²⁴ protons.
Number of protons in 34 mg NH₃ will contain
= \(6.022 \times 10^{24} \times 34 \times 10^{-3}\) = 1.2046 x 10²² Protons

(b) Mass of one proton = 1.675 x 10^-27 kg
Total mass of protons in 34 mg of NH₃
– (1.675 × 10^-27 kg) (1.2046 × 10²²)
= 2.0177 × 10^-5 kg
The number of protons, electrons and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Question 3.
How many neutrons and protons are there in the following nuclei?
\({ }_{6}^{13} \mathrm{C},{ }_{8}^{16} \mathrm{O},{ }_{12}^{24} \mathrm{Mg},{ }_{26}^{56} \mathrm{Fe},{ }_{38}^{88} \mathrm{Sr}\)
Answer:
\({ }_{6}^{13} \mathrm{C}\)C : Atomic mass =13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7
\({ }_{8}^{16} \mathrm{O}\) :
Atomic mass =16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8
\({ }_{12}^{24} \mathrm{Mg}\) : Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24-12 = 12
\({ }_{26}^{56} \mathrm{Fe}\) : Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30
\({ }_{38}^{88} \mathrm{Sr}\) : Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Question 4.
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Answer:
(i) Z = 17, A = 35
Since the no. of protons = 17 – no. of electrons
∴ The atom is chlorine, Cl: \({ }_{17}^{35} \mathrm{Cl}\)

(ii) Z = 92, A = 233
No. of protons = 92
∴ The atom is uranium, U : \({ }_{92}^{233} U\)

(iii) Z = 4, A = 9
No. of protons = 4
∴ The atom is Beryllium, Be : \({ }_{4}^{9} \mathrm{Be}\)

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (A.) of 580 nm. Calculate the frequency (υ) and wave number (\(\overline{\mathbf{v}}\)) of the ‘ yellow light.
Answer:
We know that,
Frequency, v = \(\frac{c}{\lambda}\)
1 nm = 10^-9 m
580 nm = 580 × 10^-9 m = 580 x 10^-7 cm
v = \(\) = 5.17 × 10¹⁴s^-1
(velocity of light = 3 × 10⁸ m/s)
Wave number, \(\bar{v}=\frac{1}{\lambda}\)
= \(\frac{1}{580 \times 10^{-7}}\) = 1.72 × 10⁴ cm^-1

Question 6.
Find energy of each of the photons which
(i) correspond to light of frequency 3 × 10¹⁵ Hz.
(ii) have wavelength of 0.50 Å
Answer:
(i) Energy E = hv
where, h = Planck’s constant = 6.626 × 10^-34 Js
and v = frequency of light = 3 × 10¹⁵ Hz
E= (6.626 × 10^-34)(3 × 10¹⁵)
E = 1.988 × 10^-18 J

(ii) Energy E = \(\frac{h c}{\lambda}\)
where, h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
= wavelength = 0.50Å = 0.50 × 10^-10
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{0.50 \times 10^{-10}}\) = 3.976 × 10^-15 J

Question 7.
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^-10 s.
Answer:
Frequency of light (v) = \(\frac{1}{\text { Time period }}=\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}\) = 5.0 × 10⁹s^-1
Wavelength of light (λ) = \(\frac{c}{v}\)
where, c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = \(\frac{3 \times 10^{8}}{5.0 \times 10^{9}}\) = 6.0 × 10^-2 m
Wave number(\(\bar{v}\)) of light
= \(\frac{1}{\lambda}=\frac{1}{6.0 \times 10^{-2}}\) = 1.66 × 10¹m^-1 = 16.66m^-1

Question 8.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Answer:
Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where, = wavelength of light = 4000 pm = 4000 × 10^-12 m
(∵ 1 pm = 10^-12 m)
c = velocity of light in vacuum = 3 × 10⁸ m/s
h = Planck’s constant = 6.626 × 10^-34 Js
E = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-12}}\) = 4.9695 × 10^-17J
Number of protons, N = \(\frac{1}{4.9695 \times 10^{-17}}\) = 2.0122 × 10¹⁶ protons.

Question 9.
A photon of wavelength 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10^-19 J).
Answer:
(i) Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where,
h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = wavelength of photon =4 × 10^-7 m
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 10^{-7}}\)
= 4.9695 × 10^-19j = \(\frac{4.9695 \times 10^{-19}}{1.602 \times 10^{-19}}\) [∵ 1 eV = 1.602 x 10^-19J]
= 3.10 eV
Hence, the energy of the photon is 3.10 eV.

(ii) The kinetic energy of emission E_k is given by
= hv – hv₀ = (E – W) eV
= 3.10 – 2.13 eV = 0.97 eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron is given by
KE = \(\frac{1}{2}\) mv² – hv – hv₀ = 0.97 eV
\(\frac{1}{2}\)mv² = 0.97 × 1.602 × 10^-19 J
(∵ 1 eV = 1.602 x 10^-19 J)
\(\frac{1}{2}\) × 9.11 × 10^-31 kg × v² = 0.97 × 1.602 × 10^-19 J
(∵ mass of electron = 9.11 × 10^-31 kg)
v² = \(\frac{0.97 \times 1.602 \times 10^{-19} \times 2}{9.11 \times 10^{-31}}\) = 0.341 × 10¹²
v = 0.584 × 10⁶ = 5.84 × 10⁵ m/s
Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^-1.

Question 10.
Electromagnetic radiation of wavelength 242 run is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer:
Energy (E) = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{242 \times 10^{-9} \mathrm{~m}}\)
= 0.0821 × 10^-17 J
This energy is sufficient for ionisation of one Na atom.
E = 6.02 × 10²³ × 0.0821 × 10^-17 J/mol
E = 4.945 × 10⁵ J/mol = 4.945 × 10² kJ/mol

Question 11.
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 (am. Calculate the rate of emission of quanta per second.
Answer:
Power of bulb, P = 25 watt = 25 Js^-1
Wavelength,
λ = 0.57 µm = 0.57×10^-6 m
Energy of one photon,

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800Å. Calculate threshold frequency (v₀) and work function (W₀) of the metal.
Answer:
Threshold wavelength of radiation
λ₀ = 6800Å = 6800 × 10^-10 m
Threshold frequency (v₀) of the metal
= \(\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\) = 4.41 × 10¹⁴s^-1
Thus, the threshold frequency (v₀) of the metal is 4.41 × 10¹⁴s^-1.
Hence, work function (W₀) of the metal = hv₀
= (6.626 × 10^-34Js)(4.41 × 10¹⁴s^-1)= 2.922 × 10^-19 J

Question 13.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2 ?
Answer:
The n_i = 4 to n_f= 2 transition will give rise to a spectral line of the
Balmer series. The energy involved in the transition is given by the relation,
E = 2.18 × 10^-18 \(\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)

Question 14.
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n – 1 orbit).
Answre:
Energy change, ΔE = E_f – E_i

Hence, far higher energy (25 times) is required to remove an electron from first orbit.

Question 15.
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer:
The maximum number of emission lines
= \(\frac{n(n-1)}{2}=\frac{6(6-1)}{2}\) = 3 × 5 = 15

Question 16.
(i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as :
E₅ = \(\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-2.18 \times 10^{-18}}{25}\)
E₅ = -8.72 × 10^-20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
r_n = (0.0529 nm) n²
For, n= 5
r₅ = (0.0529 nm) (5)²
r₅ = 1.3225 nm

Question 17.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
For the Balmer series, a transition from n₁= 2 to n₂ = 3 is allowed.

Question 18.
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 10^-11 ergs.
Answer:
ΔE = E₅ – E₁ = 2.18 × 10^-11 (\(\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right)\))erg
(n_i = 1st orbit and n_f = 5th orbit)

Question 19.
The electron energy in hydrogen atom is given by E_n = (-218 × 10^-18)/n²J. Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ ms^-1.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
where, λ = wavelength of moving electron
m = mass of electron = 9.1 × 10^-31 kg
h= Planck’s constant – 6.626 × 10^-34 Js
v = velocity of electron = 2.05 × 10⁷ ms^-1
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.05 \times 10^{7} \mathrm{~ms}^{-1}\right)}\)
λ = 3.548 × 10^-11 m
Hence, the wavelength of the electron moving with a velocity of
2.5× 10⁷ ms^-1 is 3.548 x 10^-11m.

Question 21.
The mass of an electron is 9.1 x 10^-31 kg. If its K.E. is 3.0 x 10^-25 J, calculate its wavelength.
Answer:
Kinetic energy, K.E. = \(\frac{1}{2}\) mv²
∴ Velocity(v) = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{2\left(3.0 \times 10^{-25}\right)}{9.1 \times 10^{-31} \mathrm{~kg}}}=\sqrt{0.6593 \times 10^{6}}\)
_ 6.626 x 10~34 Js
~ (9.1 x 10-31 kg)(811.579 ms-1)
λ = 8.968 x 10^-7 m
Hence, the wavelength of the eliectron is 8.968 x 10^-7 m or 8968 Å.

Question 22.
Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar
Ans. No. of electrons in Na⁺ = 10 [11-1]
No. of electrons in K⁺ = 18 [19-1]
No. of electrons in Mg²⁺ = 10 [12-2]
No. of electrons in Ca²⁺ =18 [20 – 2]
No. of electrons in S^2- = 18 [16 + 2]
No. of electrons in Ar = 18
∴ Na⁺ , Mg²⁺ are isoelectronic (10 e^– each)
∴ Ca²⁺, K⁺, S^2-, Ar are isoelectronic (18 e^– each).

Question 23.
(i) Write the electronic configurations of the following ions:
(a) H^– (b) Na⁺ (c)0^2- (d)F^–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s¹ (b) 2p³ (c) 3p⁵?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s²3d¹.
Answer:
(i) (a) H^– ion
The electronic configuration of H atom is 1s¹.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H^– = 1s²

(b) Na⁺ ion
The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s° or 1s² 2s² 2p⁶

(c) O^2- ion
The electronic configuration of O atom is 1s²2s² 2p⁴
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of 0^2- ion is 1s² 2s² 2p⁶

(d) F^– ion
The electronic configuration of F atom is 1s²2s²2p⁵.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of F“ ion is 1s² 2s² 2p⁶.

(ii) (a) 3s¹
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s1. .-. Number of electrons present in the atom of the element = 2 +2 + 6 +1=11
∴ Atomic number of the element = 11

(b) 2p³
Completing the electronic configuration of the element as 1s² 2s² 2p³
∴ Number of electrons present in the atom of the element = 2+ 2+ 3 = 7
∴ Atomic number of the element = 7

(c) 3p⁵
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s² 3 P⁵
∴ Number of electrons present in the atom of the element
2 + 2 + 6 + 2 + 5 = 17
∴ Atomic number of the element = 17

(iii) (a) [He] 2s¹
The electronic configuration of the element is [He]2s¹ = 1s²2s¹.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³
The electronic configuration of the element is
[Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹
The electronic configuration of the element is [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s² 3d² is scandium (Sc).

Question 24.
What is the lowest value of n that allows g orbitals to exist?
Answer:
For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n —1).
∴ For l = 4, minimum value of n = 5

Question 25.
An electron is in one of the 3d orbitais. Give the possible values of n, l and m_l for this electron.
Answer:
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (L) = 2
Magnetic quantum number (m_l) = – l to + l including 0 = -2 -1, 0, 1, 2

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers ?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

Question 33.
Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.
Answer:

Question 34.
Calculate the energy required for the process :
He⁺fe) → He²⁺(g) + e^–
The ionisation energy’ for the H atom in the ground state is 2.18 × 10^-18  J atom^-1
Answer:

For H atom (Z = 1), E_n =2.18 × 10^-18 × (l)² J atom^-1 (given)
For He⁺ ion (Z = 2), E_n =2.18 × 10^-18 × (2)² = 8.72 × 10^-18 J atom^-1 (one electron species)

Question .35.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Answer:

Question 36.
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Answer:
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

Radius of each carbon atom = 12(1.2 × 10^-8) = 6.0 × 10^-9cm = 0.06 nm

Question 37.
The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
Answer:
(a) Radius of zinc atom =Undefined control sequence = 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

Question 38.
A certain particle carries 2.5 x 10^-16 C of static electric charge. Calculate the number of electrons present in it.
Answer:

Question 39.
In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10^-18C. Calculate the number of electrons present in it.
Answer:

Question 40.
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Answer:
We have studied that in the Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible.

Question 41.
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br can be written whereas symbols \(_{ 79 }^{ 35 }{ Br }\) and ₃₅Br are not accepted. Answer in brief.
Answer:
In the symbol \(_{ A }^{ B }{ X }\) of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol \(_{ 79 }^{ 35 }{ Br }\) is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [\(\frac { x\times 31.7 }{ 100 } \) = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = \(\frac { 81 }{ 2.317 } \) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine \(_{ 35 }^{ 81 }{ Br }\).

Question 43.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
Answer:
Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + \(\frac { x\times 11.1 }{ 100 } \) = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = \(\frac { 38 }{ 2.111 } \) = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = \(_{ 17 }^{ 37 }{ Cl }\)^–

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac { x\times 31.7 }{ 100 } \) = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = \(\frac { 53 }{ 2.304 } \) = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 45.
Arrange the following type of radiations in increasing order of frequency:
(a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
Answer:
The increasing order of frequency is as follows :
Radiation from FM radio < Radiation from microwave oven < Amber light from traffic signal < X-rays < Cosmic rays from outer space.

Question 46.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴ , calculate the power of this laser.
Answer:
Power of laser (E) = \(\frac{n h c}{\lambda}\)
= \(\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{~m}\right)}\)
= 0.3302 x 10⁷ J = 3.33 x 10⁶ J
Hence, the power of the laser is 3.33 x 10⁶ J.

Question 47.
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Answer:
(a) Frequency (v) of emission can be calculated as: c = v x λ
where, λ – wavelength = 616 nm = 616 x 10^-9 ms^-1
c = velocity of light = 3.0 x 10⁸ ms^-1
v = \(\frac{c}{\lambda}=\frac{3.0 \times 10^{8}}{616 \times 10^{-9}}\) = 4.87 x 10¹⁴s^-1

(b) Distance travelled in 1 second = 3.0 x 10⁸ m
Distance travelled in 30 seconds = 30 x 3.0 x 10⁸ m = 9 x 10⁹ m
(c) Energy of the photon (quantum)
E= hv = 6.626 x 10^-34 x 4.87 x 10¹⁴J
= 32.26 x 10^-20 J

(d) Number of quanta = \(\frac{Total energy produced}
{Energy of one quantum}\)
= \(\frac{2}{32.26 \times 10^{-20}}\) = 6.2 x 10¹⁸

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 rnn, calculate the number of photons received by the detector.
Answer:
Energy of one photon, E = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}\) = 3.313 x 10^-19 J
Number of photons = \(\frac{Total energy received}{Energy of one photon}\)
= \(\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}\)

Question 49.
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.
Answer:
Duration of radiation source = 2 ns = 2 x 10^-9 s
No. of photons (n) = 2.5 x 10¹⁵
Frequency of radiation
v = \(\frac{1}{2.0 \times 10^{-9} \mathrm{~s}}\) = 5.0 x 10⁸s^-1
Energy (E) of source = n x hv
E= (2.5 x 10¹⁵)(6.626 x 10^-34Js) (5.0 x 10⁸)
E = 8.282 x 10^-10J
Hence, the energy of the source (E) is 8.282 x 10^-10 J.

Question 50.
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
For λ₁ = 589 nm = 589 x 10^-9
Frequency of transition (v₁) = \(\) = 5.093 x 10¹⁴s^-1
Similarly, for λ₂ = 589.6 nm = 589.6 x 10^-9m
Frequency of transition (v₂) = \(\frac{c}{\lambda_{2}}=\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{589.6 \times 10^{-9} \mathrm{~m}}\) = 5.088 x 10¹⁴s^-1
Energy difference (ΔE) between excited states = E₁ – E₂
ΔE = h(v₁ – v₂)
= (6.626 × 10^-34Js) (5.093 × 10¹⁴ – 5.088 × 10¹⁴)s^-1
= (6.626 × 10^-34Js) (0.005 × 10^-14) s^-1
ΔE = 3.31 × 10^-22

Question 51.
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer:
It is given that the work function (W₀) for caesium atom is 1.9 eV.
(a) Work function,

λ₀ = 6.53 × 10^-7 m
= 653 × 10^-9m = 653nm.
Hence the threshold wave length is 653nm

(b) threshold frequency
v₀ = \(\frac{W_{0}}{h}\)
v₀ = \(\frac{1.9 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\) = 4.59 x 10¹⁴s^-1
Hence the threshold frequency of radiation is 4.59 x 10¹⁴s^-1
According to the question :
wavelength used in irradiation = 500nm
Kinetic Energy = h(v – v₀)
= hc( \(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\) )

Question 52.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

Answer:

Mass of electron, m = 9.1 × 10^-31 kg
Here, λ = 500 × 10^-9 m
v_max = 2.55 ×10⁵ms^-1
λ₀ = ?

Question 53.
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:
h = 6.626 × 10^-34 Js
V₀ = 0.35 volt
e = 1.6 × 10^-19 C
λ = 256.7 nm = 256.7 × 10^-9 m
W₀ = ? (eV)

Question 54.
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ms^-1, calculate the energy with which it is bound to the nucleus.
Answer:
λ = wavelength = 150 pm = 150 × 10^-12 m
v = velocity = 1.5 × 10⁷ ms^-1
Kinetic energy of the ejected electron
= \(\frac{1}{2}\) mv⁷ = \(\frac{1}{2}\) x 9.1 x 10^-31 x (1.5 x 10⁷ )² J = 0.102 x 10^-15 J
Energy of the incident radiation
hv= hv₀ +\(\frac{1}{2}\)mv²
E = hv = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{150 \times 10^{-12}}\)
= 1.325 × 10^-15 J
Minimum energy required to eject the electron
E₀ = E – K.E.
= (1.325 × 10^-15 – 0.102 × ^-15 ) J
= 1.223 × 10^-15J
= \(\frac{1.223 \times 10^{-15}}{1.602 \times 10^{-19}} \mathrm{eV}\)
[ 1J = 1.602 × 10^-19 eV]
= 7.6 × 10³ eV

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 10¹⁵ (Hz) \(\left[\frac{1}{3^{2}}-\frac{1}{n^{2}}\right]\)
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

Question 56.
Calculate the wavelength for the emis8ion transition if it starts from the orbit having radius 1.3225 mn and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
The radius of the nth orbit of hydrogen like particles is given by

Thus, the transition ocurs from the 5th orbit to the 2nd orbit. It belongs to the Balmer series.
Wave number ( \(\bar{V}\) ) for the transition is given by,
\(\bar{v}=\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)=1.097 \times 10^{7}\left(\frac{21}{100}\right)\)
= 2.303 × 10⁶ m^-1
∴ Wavelength ) associated with the emission transition is given by,
λ = \(\frac{1}{\bar{v}}=\frac{1}{2.303 \times 10^{6} \mathrm{~m}^{-1}}\)
= 0.434 × 10^-6 = 434 nm
∴ This transition belongs to Balmer series and comes in the visible region of the spectrum.

Question 57.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ ms^-1, calculate de Broglie wavelength associated with this electron.
Answer:
From Broglie’s equation,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(1.6 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) (1J = 1 kgm²s^-2)
= 4.55 × 10^-10 m = 455 pm
∴ de Broglie’s wavelength associated with the electron is 455 pm.

Question 58.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:
From de Broglie’s equation,
λ = \(\frac{h}{m v}\)
v = \(\frac{h}{m \lambda}\)
mass of neutron, m = 1.675 × 10^-27 kg
λ = 800pm = 800 × 10^-12 m
v = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(1.675 \times 10^{-27} \mathrm{~kg}\right)\left(800 \times 10^{-12} \mathrm{~m}\right)}\) = 4.94 × 10²ms^-1
= 494 ms^-1
∴ Velocity associated with the neutron is 494 ms^-1.

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ms^-1, calculate the de Broglie wavelength associated with it.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
Mass of electron = 9.1 × 10^-31 kg, h = 6.626 × 10^-34 Js,
velocity = 2.19 × 10⁶ ms^-1
= \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) = 3.32 × 10^-10m
= 3.32 × 10^-10m × \(\frac{100}{100}\) = 332 × 10^-12 m
λ = 332pm
∴ Wavelength associated with the electron is 332 pm.

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 437 × 10⁵ ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:
Wavelength associated with the velocity of hockey ball,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{(0.1 \mathrm{~kg})\left(4.37 \times 10^{5} \mathrm{~ms}^{-1}\right)}\)
λ = 1.516 × 10^-38 m

Question 61.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac{h}{4 \pi_{m}}\) × 0.05 nm, is there any problem in defining this value.
Answer:
From Heisenberg’s uncertainty principle,

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these comhination(s) has/have the same energy lists.
1. n = 4, l = 2, m_l = -2, m_s = – \(\frac{1}{2}\)
2. n = 3, l = 2, m_l= 1, m_s = + \(\frac{1}{2}\)
3. n = 4, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
4. n = 3, l – 2, m_l = -2, m_s = –\(\frac{1}{2}\)
5. n = 3, l = 1, m_l = -1, m_s = + \(\frac{1}{2}\)
6. n = 4, l = 1, m_l = 0, m_s = + \(\frac{1}{2}\)
Answer:
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5 and 6 are present in the 4d, 3d, 4p, 3d, 3p and 4p orbitals respectively.
Therefore, the increasing order of energies are
5(3p) < 2(3d) = 4(3d) < 6(4p) = 3(4p) < 1(4d).

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital,6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge.
Answer:
Effective nuclear charge decreases as the distance of the orbitals increases from the nucleus. Hence, 4p electrons experience the lowest effective nuclear charge.

Question 64.
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
Answer:
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4/ since 4d is closer to the nucleus.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3/.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer:
₁₃Al = 1s²2s²2p⁶3s²3p¹
₁₄Si = 1s²2s²2p⁶3s²3p²
Si (+4) has greater nuclear charge than aluminium (+3). Hence, 3p unpaired electrons of Si experience greater effective nuclear charge than Al.

Question 66.
Indicate the number of unpaired electrons in
(a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer:
(a) Phosphorus (P) : Atomic number = 15
The electronic configuration of P is : 1s² 2s² 2p⁶ 3s² 3p³

The orbital picture of P can be represented as :

From the orbital picture, phosphorus has three unpaired electrons
(b) Silicon (Si) : Atomic number = 14
The electronic configuration of Si is : 1s² 2s² 2p⁶ 3s² 3p²
The orbitai picture of Si can be represented as :

From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr) : Atomic number = 24
The electronic configuration of Cr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
The orbital picture of chromium is :

From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe) : Atomic number = 26

The electronic configuration of Fe is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
The orbital picture of Fe is:

From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr) : Atomic number = 36
The electronic configuration of Kr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
The orbital picture of krypton is :

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Question 67.
(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m_s value of \(\) 1/2 for n – 4?
Answer:
(a) n = 4 (Given)
For a given value of ‘rt, V can have values from zero to n -1.
l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f
(b) Number of orbitals in the nth shell = n²
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with m_s value of \(-\frac{1}{2}\)
Number of electrons with ms value of (\(-\frac{1}{2}\)) is 16.