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

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On April 24, 2024, 11:35 AM

Question 12.11:

Answer the following Questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

(a) Is the average angle of deflection of α­-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?

Answer:

(a) about the same

The average angle of deflection of α­-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

(b) much less

The probability of scattering of α-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

(c) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

(d) Thomson’s model

It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of α­−particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Question 12.12:

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10⁻⁴⁰. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the Answer interesting.

Answer:

Radius of the first Bohr orbit is given by the relation,

Where,

∈₀ = Permittivity of free space

h = Planck’s constant = 6.63 × 10⁻³⁴ Js

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

e = Charge of an electron = 1.9 × 10⁻¹⁹ C

m_p = Mass of a proton = 1.67 × 10⁻²⁷ kg

r = Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as: 

Gravitational force of attraction between an electron and a proton is given as:

Where,

G = Gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

∴F_G = F_C

Putting the value of equation (4) in equation (1), we get:

It is known that the universe is 156 billion light years wide or 1.5 × 10²⁷ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Question 12.13:

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n−1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Answer:

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n−1).

We have the relation for energy (E₁) of radiation at level n as:

Now, the relation for energy (E₂) of radiation at level (n − 1) is givenas: 

Energy (E) released as a result of de-excitation:

E = E₂−E₁

hν = E₂ − E₁ … (iii)

Where,

ν = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

For large n, we can write

Classical relation of frequency of revolution of an electron is given as: 

Where,

Velocity of the electron in the n^th orbit is given as:

v =

And, radius of the n^th orbit is given as:

r = 

Putting the values of equations (vi) and (vii) in equation (v), we get:

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.