AIEEE 2010 Questions (MCQ) from Nuclear Physics

“I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend”

– Sir Isaac Newton

Multiple choice questions based on a given paragraph are seen in many entrance exam question papers. Here are two such questions on nuclear physics (which appeared in AIEEE 2010 question paper):

Directions : Questions (i) and (ii) are based on the following paragraph:

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

(i) The speed of daughter nucleus is

(1) c√[Δm/(M + Δm)]

(2) c [Δm/(M + Δm)]

(3) c√(2Δm/M)

(4) c√(Δm/M)

The energy released in the process is Δmc² in accordance with Einstein’s mass-energy relation. This energy is carried by the daughter nuclei (as kinetic energy).

Therefore we have

½ (M/2)v² + ½ (M/2)v² = Δmc² where v is the speed of each daughter nucleus.

Or, Mv²/2 = Δmc² from which v = c√(2Δm/M), as given in option (3).

[Note that the momentum is to be conserved in the process. Therefore, the daughter nuclei have to travel in opposite directions with the same speed since they have the same mass. If you were asked to give the velocities of the daughters, the answer would be v and – v]

(ii) The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then

(1) E₁ = 2 E₂

(2) E₂ = 2 E₁

(3) E₁ > E₂

(4) E₂ > E₁

The binding energy per nucleon for the daughter nuclei is greater than that for the parent nucleus since energy is released in the process[Option (4)].

The following question (MCQ) also was included in the AIEEE 2010 question paper:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1) (A – Z – 4)/(Z – 2)

(2) (A – Z – 8)/(Z – 4)

(3) (A – Z – 4)/(Z – 8)

(4) (A – Z – 12)/(Z – 4)

When an α-particle is emitted, the nucleus loses two protons and two neutrons. When a positron is emitted by the nucleus, a proton in the nucleus gets converted into a neutron. Since the number of neutrons in the original neucleus is A – Z, the number of neutrons in the final nucleus will be (A – Z) – (3×2) + 2 = A – Z – 4.

The protons in the final nucleus will be Z – (3×2) – 2 = Z – 8

Therefore, the ratio of number of neutrons to that of protons in the final nucleus will be(A – Z – 4)/(Z – 8), as given in option (3).