Atoms Question Answers: NCERT Class 12 Physics

(a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.

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

(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.

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.

Rydberg’s formula is given as: hc/λ = 21.76 x 10^-19 [1/(n₁)² - 1/(n₂)²]

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

c = Speed of light = 3 × 10 ⁸ m/s

(n ₁ and n ₂ are integers)

The shortest wavelength present in the Paschen series of the spectral lines is given for values n ₁ = 3 and n ₂ = ∞.

hc/λ = 21.76 x 10^-19 [1/(3)² - 1/(∞)²]

∴ λ = 6.6 x 10^-34 x 3 x 10⁸ x 9 / 21.76 x 10^-19 = 8.189x10^-7 m = 818.9 nm

Hence, the shortest wavelength present is 818.9 nm.

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 = 6.62 x 10^-34 Js

∴ V = E/h = 3.68 x 10^-19 / 6.62 x 10^-32 = 5.55 x 10¹⁴ Hz

Hence, the frequency of the radiation is 5.6 × 10 ¹⁴ Hz.

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

For ground level, n ₁ = 1

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

E₁ = -13.6/n₁² eV

= -13.6/1² = -13.6 eV

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

Let E² be the energy of this level.

∴ E₂ = -13.6/n₂² eV

= -13.6/4² = -13.6/16 eV

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

E = E₂ - E₁

= (-13.6 /16) - (-13.6/1)

= 13.6 X 15/16 eV

= (13.6 X 15/16) X 1.6 X 10^-19 = 2.04 X 10^-18 J

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

E = hc/λ

Where,

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

c = Speed of light = 3 × 10⁸ m/s

∴ λ = hc/E

= (6.6x10^-34x3x10⁸)/(2.04x10^-18)

= 9.7x10^-8 m = 97 nm

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

v = c/λ

= (3x10⁸)/(9.7x10^-8) ≈ 3.1 x 10¹⁵ Hz

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

(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,

ν ₁ = e²/n₁4πϵ₀(h/2π) = e²/2ϵ₀h

Where, e = 1.6 × 10⁻¹⁹ C

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

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

∴ ν₁ = (1.6x10^-19)2/2x8.85x10^-12x6.62x10^-34 = 0.0218 x 10⁸ = 2.18 x 10⁶ m/s

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

ν₂ = e²/n₂2ϵ₀h = (1.6x10^-19)2/2x2x8.85x10^-12x6.62x10^-34 = 1.09 x 10⁶ m/s

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

ν₃ = e²/n₃2ϵ₀h = (1.6x10^-19)2/3x2x8.85x10^-12x6.62x10^-34 = 7.27 x 10⁵ m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10 ⁶ m/s, 1.09 × 10 ⁶ m/s, 7.27 × 10 ⁵ m/s respectively.

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

Orbital period is related to orbital speed as:

T₁ = 2πr₁/ν ₁

Where, r₁ = Radius of the orbit

= n₁²h²ϵ₀/πme²

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

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

∴ T₁ = 2πr₁/ν ₁

= (2πx(1)²x(6.62x10^-34)2x8.85x10^-12)/2.18x10⁶xπx9.1x10^-31x(1.6x10^-19)²

= 15.27x10^-17 = 1.527x10^-16 s

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

T₂ = 2πr₂/ν ₂

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

= (n₂)²h²ϵ₀/πme²

∴ T₂ = 2πr₂/ν₂

= (2πx(2)²x(6.62x10^-34)²x8.85x10^-12)/1.09 x 10⁶ x π x 9.1 x 10^-31 x (1.6 x 10^-19)²

= 1.22 x 10^-15 s

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

T₃ = 2πr₃/ν ₃

Where, r ₃ = Radius of the electron in n ₃ = 3

= (n₃)²h²ϵ₀/πme²

∴ T₃ = 2πr₃/ν ₃

= (2πx(3)²x(6.62x10^-34)2x8.85x10^-12)/7.27 x 10⁵ x π x 9.1 x 10^-31 x (1.6 x 10^-19)²

= 4.12 x 10^-15 s

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.

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:

r₂ = (n)²r₁

= 4 x 5.3 x 10^-11 = 2.12 x 10^-10 m

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

r₃ = (n)² r₁

= 9 x 5.3 x 10^-11 = 4.77 x 10^-10 m

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.

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 = -13.6/(n)² eV

For n = 3, E = -13.6/(9)² = -1.5 eV

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:

1/λ = R_y (1/1² - 1/n²)

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:

1/λ = 1.097 x 10⁷ (1/1² - 1/3²)

= 1.097x10⁷(1-1/9) = 1.097x10⁷x8/9

λ = 9/(8 x 1.097 x 10⁷) = 102.55 nm

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

1/λ = 1.097 x 10⁷ (1/1² - 1/2²)

= 1.097x10⁷(1-1/4) = 1.097x10⁷x3/4

λ = 4/(1.097x10⁷x3) = 121.54nm

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

1/λ = 1.097 x 10⁷ (1/2² - 1/3²)

= 1.097x10⁷(1/4-1/9) = 1.097x10⁷x5/36

λ = 36/(5x1.097x10⁷) = 656.33 nm

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.

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:

mvr = nh/2π

Where,

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

n = Quantum number

∴ n = mvr2π/h

= (2πx6x10²⁴x3x10⁴x1.5x10¹¹)/(6.62x10^-34)

= 25.61x10⁷³ = 2.6 x 10⁷⁴

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