AP Physics B – Some Interesting Multiple Choice Practice Questions on Atomic and Nuclear Physics

“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages.”

– Thomas A. Edison

The sections ‘atomic physics and quantum effects’ and ‘nuclear physics’ in the AP Physics B syllabus will appear to be interesting to most of the AP Physics B aspirants. Today we will discuss a few practice questions (multiple choice) in these sections:

(1) The de Broglie wave length of a particle with kinetic energy E is λ. If its kinetic energy is increased  to 4E, its de Broglie wave length will be

(a) λ/4

(b) λ/2

(c) λ

(d) 2λ

(e) 4λ

Since the kinetic energy is directly proportional to the square of momentum, the momentum of the particle is doubled when its kinetic energy is quadrupled.

[Note that kinetic energy E = p²/2m where p is the momentum and m is the mass].

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Therefore, when p is doubled, λ is halved [Option (b)].

(2) Particles A and B have masses m and 4m respectively but they carry the same charge. When they are accelerated by the same voltage, their de Broglie wave lengths are in the ratio

(a) 1 : 1

(b) 2 : 1

(c) 4 : 1

(d) 1 : 4

(e) 1 : 8

Let V represent the common accelerating voltage and q represent the common charge of the particles. If p₁ and p₂ are the momenta of the particles A and B, on equating their kinetic energies, we have

             p₁²/2m = p₂²/(2×4m)

[Remember that the kinetic energy of a particle of charge q accelerated by a voltage V is qV. The kinetic energies of A and B are equal since they have the same charge and they are accelerated by the same voltage]

The above equation gives

             p₁/ p₂ = ½

Since the de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum, the ratio of the de Broglie wave lengths of A and B is given by

              λ₁/ λ₂ = p₂/ p₁ = 2, as given in option (b).

(3) A metallic surface is found to emit photo-electrons when monochromatic light rays of frequencies n₁ and n₂ (n₂ > n₁) are incident on it. If the maximum values of kinetic energy of the photo-electrons emitted in the two cases are in the ratio 1 : 3, the threshold frequency of the metallic surface is

(a) (3n₁ – n₂)/ 2

(b) (2n₁ – n₂)/ 2

(c) (3n₁ – n₂)/ 3

(d) (n₂ – n₁)/ 3

(e) (n₂ – n₁)/ 2

If the threshold frequency is n₀ and the maximum values of kinetic energy in the two cases are E₁ and E₂ respectively, we have

             hn₁ = hn₀ + E₁ and

             hn₂ = hn₀ + E₂

Therefore, E₁/E₂ = (n₁ – n₀)/(n₂ – n₀)

Since the ratio is 1 : 3 we have

             (n₁ – n₀)/(n₂ – n₀) = 1/3

Or, 3n₁ – 3n₀ = n₂ – n₀

This gives n₀ = (3n₁ – n₂)/ 2, as given in option (a).

(4) The energy that must be added to an electron to reduce its de Broglie wave length from 2 nm to 1 nm is

(a) half the initial energy

(b) equal to the initial energy

(c) twice the initial energy

(d) thrice the initial energy

(e) four times the initial energy

The de Broglie wave length λ is given by

             λ = h/p where h is Planck’s constant and p is the momentum.

Since the de Broglie wave length of the electron is to be reduced to half the initial value (from 2 nm to 1 nm), the momentum of the electron is to be doubled. But when the momentum is doubled, its kinetic energy becomes four times the initial value. Therefore the energy that must be added is three times the initial energy [Option (d)].

(5) An alpha particle of mass m and speed v  proceeds directly towards a heavy nucleus of charge Ze. The distance of closest approach of the alpha particle is directly proportional to

(a) 1/Ze²

(b) 1/Ze

(c) v

(d) m

(e) 1/v²

The alpha particle has to move towards the nucleus with difficulty, doing work against the electrostatic repulsive force. When the alpha particle reaches the distance of closest approach, the entire kinetic energy gets converted into electrostatic potential energy. Therefore we have

             ½ mv² = (1/4πε₀)(2Ze²/r) where r is the distance of closest approach.

[Remember that the charge on the alpha particle is 2e].

This gives r = Ze²/πε₀mv²

This shows that r is directly proportional to 1/v².