Ncert Solution for Class 12 Physics Chapter 12 Atoms

Atoms ncert solutions: Ncert Class 12 Physics Chapter 12 Exercise Solutions

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Question 12.1: Choose the correct alternative from the clues given at the end of the each statement:
(a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.)

Solution 12.1:

(a)  The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

(b) In the ground state of Thomson’s model, electrons are in stable equilibrium, while in Rutherford’s model, electrons always experience a net force.

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(c) A classical atom based on Rutherford’s model is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model.

(e) The positively charged part of the atom possesses most of the mass in both the models.

Question 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Solution 12.2: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10⁻²⁷ kg) is less than the mass of incident α−particles (6.64 × 10⁻²⁷ kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

Question 12.3: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Solution 12.3: It is given that

Separation of two energy levels in an atom,

• E = 2.3 eV
• = 2.3 × 1.6 × 10^-19
• = 3.68 × 10^-19 J

Consider v as the frequency of radiation emitted when the atom transits from the upper level to the lower level.

So, the relation for energy can be written as: E = hv

Where,

h = Planck’s constant = 6.62 × 10^-34 Js

v = E/h

= \(\frac{3.68 \times 10^{-19}}{6.62\times 10^{-32}}\)

= 5.55 × 10¹⁴ Hz

Hence, the frequency of the radiation is 5.6 × 10¹⁴ Hz.

Question 12.4: The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Solution 12.4: In a hydrogen atom, the total energy ( E ) of the electron in the ground state is given by the formula:

E = K + U

where ( K ) is the kinetic energy and ( U ) is the potential energy. For a hydrogen atom, the relationships between these energies are as follows:

1. The potential energy ( U ) of the electron in the ground state is given by: U = -2K
2. The total energy ( E ) can be expressed as: E = K + U = K – 2K = -K

Given that the ground state energy of the hydrogen atom is E = -13.6 eV:

\(-K = -13.6 \, \text{eV} \implies K = 13.6 \, \text{eV}\)

Now, using the relationship U = -2K:

\(U = -2 \times 13.6 \, \text{eV} = -27.2 \, \text{eV}\)

In summary:

• Kinetic energy K = 13.6 eV
• Potential energy U = -27.2 eV

Question 12.5: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon

Solution 12.5: For ground level, n₁= 1

Let E₁ be the energy of this level. It is known that E₁ is related to n₁ as:

\(E_{1}=\frac{-13.6}{{(n_{1})^{2}}}\)eV

\(=\frac{-13.6}{1^{2}}\)

= -13.6 eV

The atom is excited to a higher level, n₂= 4

Let E₂ be the energy of this level: \(E_{2}=\frac{-13.6}{{(n_{2})^{2}}}\)eV

\(=\frac{-13.6}{4^{2}}\)

\(=\frac{-13.6}{16}\) eV

Following is the amount of energy absorbed by the photon: E = E₂− E₁

= \(\frac{-13.6}{16}-(\frac{-13.6}{1})\)

= \(\frac{-13.6\times 15}{16}\)eV

= \(\frac{-13.6\times 15}{16}\times 1.6\times 10^{-19}\)

= 2.04 × 10^-18 J

For a photon of wavelength λ, the expression of energy is written as: \(E = \frac{hc}{\lambda }\)

Where,

h = Planck’s constant = 6.6 × 10⁻³⁴ Js

c = Speed of light = 3 × 10⁸ m/s

\(\lambda = \frac{hc}{E}\)

\(\ = \frac{6.6 \times 10 ^{-34}\times 3\times 10^{8}}{2.04 \times 10^{-18}} \)

= 9.7 × 10^-8 m

= 97nm

And, frequency of a photon is given by the relation, v = c/λ

\(\frac{3\times 10^{8}}{9.7\times 10^{-8}}\)

= 3.1 × 10¹⁵ Hz

Therefore, 97 nm and 3.1×10¹⁵ Hz is the wavelength and frequency of the photon.

Question 12.6: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Solution 12.6: (a) Let v₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁= 1. For charge (e) of an electron, v₁ is given by the relation,

\(v_1 = \frac{e^{2}}{n_1 4\;\pi\;\epsilon_0\; (\frac{h}{2\pi}) } = \frac{e^{2}}{2\;\epsilon_0 \;h}\)

Where, e = 1.6 × 10⁻¹⁹ C

ϵo = permittivity of free space. = 8.85 × 10⁻¹² N⁻¹ C ² m⁻²

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

\(v_1 = \frac{(1.6\times10^{-19})^{2}}{2\times8.85\times10^{-12}\times6.62\times10^{-34}}\)

= 0.0218 x 10⁸ 

= 2.18 × 10⁶m/s

For level n₂= 2, we can write the relation for the corresponding orbital speed as:

\(v_2 = \frac{e^2}{n_2 2 \epsilon_0 h}\)

\(= \frac{(1.6 \times 10^{-19})^2}{2 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}\)

= 1.09 × 10⁶ m/s

For level n₃= 3, we can write the relation for the corresponding orbital speed as:

\(v_3 = \frac{e^2}{n_3 2 \epsilon_0 h}\)

\(= \frac{(1.6 \times 10^{-19})^2}{2 \times 3 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}\)

= 7.27 × 10⁵ m/s

Therefore, in a hydrogen atom, the speed of the electron at different levels, that is, n = 1, n = 2, and n = 3, is 2.18×10⁶m/s, 1.09×10⁶m/s and 7.27 × 10⁵ m/s.

(b) Let T₁ be the orbital period of the electron when it is in level n₁= 1.

The orbital period is related to orbital speed as:

\(T_1 = \frac{ 2 \pi r_1 } { v_1 }\)

 [Where, r₁ = Radius of the orbit]

\(=\frac{ { n _ 1 } ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

e = Charge on an electron = 1.6 × 10⁻¹⁹ C

ε₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

m = Mass of an electron = 9.1 × 10⁻³¹ kg

For level n₂ = 2, we can write the period as:

\(T_2 = \frac{ 2 \pi r_2 } { v_2 }\)

[Where, r₂ = Radius of the electron in n₂ = 2]

\(=\frac{ {( n _ 2 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

And, for level n3 = 3, we can write the period as:

\(T_3= \frac{ 2 \pi r_3 } { v_3 }\)

[Where, r₃ = Radius of the electron in n₃ = 3]

\(=\frac{ {( n _ 3 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)

Therefore, 1.52 × 10^-16s, 1.22 × 10^-15s and 4.12 × 10^-15s are the orbital periods in each level.

Question 12.7: The radius of the innermost electron orbit of a hydrogen atom is 5.3×10^–11 m. What are the radii of the n = 2 and n =3 orbits?

Solution 12.7: The radius of the innermost orbit of a hydrogen atom, r₁ = 5.3 × 10⁻¹¹ m.

Let r₂ be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as: \(r_2 = ( n ) ^ { 2 } r_ 1\)

= 4 × 5.3 × 10^-11 = 2.12 × 10^-10 m

For n = 3, we can write the corresponding electron radius as: \(r_3 = ( n ) ^ { 2 } r_ 1\)

= 9 × 5.3 x 10^-11 = 4.77 × 10^-10 m

Therefore, 2.12 × 10⁻¹⁰ m and 4.77 × 10⁻¹⁰ m are the radii of an electron for n = 2 and n = 3 orbits, respectively.

Question 12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution 12.8: It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV, i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as: \(E = \frac {-13.6 } { n ^ { 2 } } \; eV\)

E = -13.6 / 9 = -1.5 eV

For n = 3,

This energy is approximately equal to the energy of gaseous hydrogen.

It can be concluded that the electron has jumped from n = 1 to the n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for the Lyman series as: \(\frac { 1 } { \lambda } = R_y \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { n ^ { 2 } } \right )\)

Where, R_y= Rydberg constant = 1.097 × 10⁷ m⁻¹

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 9 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 8 } { 9 } \right )\)

\(\lambda = \frac { 9 }{8 \times 1.097 \times 10 ^ {7 }}\)

= 102.55 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 2 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 4 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 3 } { 4 } \right )\)

\(\lambda = \frac { 4 }{3 \times 1.097 \times 10 ^ {7 }}\)

= 121.54 nm

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 2 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 4 } -\frac { 1 } { 9 } \right )\)

\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 5 } { 36 } \right )\)

\(\lambda = \frac { 36 }{ 5 \times 1.097 \times 10 ^ {7 }}\)

= 656.33 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Therefore, there are two wavelengths that are emitted in the Lyman series, which are approximately 102.5 nm and 121.5 nm, and one wavelength in the Balmer series, which is 656.33 nm.

Question 12.9: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)

Solution 12.9: he radius of the orbit of the Earth around the Sun, r = 1.5 × 10¹¹ m

The orbital speed of the Earth, ν = 3 × 10⁴ m/s

Mass of the Earth, m = 6.0 × 10²⁴ kg

According to Bohr’s model, angular momentum is quantised and given as:

\(m v r = \frac { n h } { 2 \pi }\)

Where,

h = Planck’s constant = 6.62 × 10⁻³⁴ J s

n = Quantum number

\(n = \frac { m v r 2 \pi } { h }\)

\(= \frac{ 2 \pi \times 6 \times 10^{24} \times 3 \times 10 ^ {4} \times 1.5 \times 10^{11} }{ 6.62 \times 10^{-34} }\)

= 25.61 ×10⁷³ = 2.6 × 10⁷⁴

Therefore, the Earth revolution can be characterised by the quantum number 2.6 × 10⁷⁴.