Page No 436: - Chapter 12 Atoms Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 12.3:

What is the shortest wavelength present in the Paschen series of spectral lines?

Answer:

Rydberg’s formula is given as:

Where,

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

c = Speed of light = 3 × 10⁸ m/s

(n₁ and n₂ are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁ = 3 and n₂ = ∞.

Question 12.4:

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

Answer:

Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10⁻¹⁹

= 3.68 × 10⁻¹⁹ J

Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

We have the relation for energy as:

E = hv

Where,

h = Planck’s constant

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

Question 12.5:

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

Answer:

Ground state energy of hydrogen atom, E = − 13.6 eV

This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.

Kinetic energy = − E = − (− 13.6) = 13.6 eV

Potential energy is equal to the negative of two times of kinetic energy.

Potential energy = − 2 × (13.6) = − 27 .2 eV

Question 12.6:

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 the photon.

Answer:

For ground level, n₁ = 1

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

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

Let E₂ be the energy of this level.

The amount of energy absorbed by the photon is given as:

E = E₂ − E₁

For a photon of wavelengthλ, the expression of energy is written as:

Where,

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

c = Speed of light = 3 × 10⁸ m/s

And, frequency of a photon is given by the relation,

Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 10¹⁵ Hz.

Question 12.7:

(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.

Answer:

(a) Let ν₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 1. For charge (e) of an electron, ν₁ is given by the relation,

Where,

e = 1.6 × 10⁻¹⁹ C

∈₀ = Permittivity of free space = 8.85 × 10⁻¹² N⁻¹ C² m⁻²

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

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

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

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10⁶ m/s, 1.09 × 10⁶ m/s, 7.27 × 10⁵ m/s respectively.

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

Orbital period is related to orbital speed as:

Where,

r₁ = Radius of the orbit

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

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

∈₀ = 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:

Where,

r₂ = Radius of the electron in n₂ = 2

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

Where,

r₃ = Radius of the electron in n₃ = 3

Hence, the orbital period in each of these levels is 1.52 × 10⁻¹⁶ s, 1.22 × 10⁻¹⁵ s, and 4.12 × 10⁻¹⁵ s respectively.

Question 12.8:

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?

Answer:

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:

For n = 3, we can write the corresponding electron radius as:

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

Question 12.9:

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

Answer:

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:

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 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 Lyman series as:

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 λas:

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

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

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

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

Question 12.10:

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.)

Answer:

Radius of the orbit of the Earth around the Sun, r = 1.5 × 10¹¹ m

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 quantized and given as:

Where,

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

n = Quantum number

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 10⁷⁴.