All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions on Nuclear Physics

Everything should be made as simple as possible, but not simpler

– Albert Einstein

The following questions on nuclear physics appeared in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper:

1. In the nuclear decay given below:

^AX_Z → ^AY_Z+1→ ^{A– 4} B* _{Z– 1}→ ^{A– 4} B _{Z– 1},

the particles emitted in the sequence are

(1) γ, β, α

(2) β, γ, α

(3) α, β, γ

(4) β, α, γ

The nucleus ^AX_Z emits a β-particle and its atomic number increases by 1 to transform to the nucleus ^AY_Z+1. The nucleus ^AY_Z+1 emits an α-particle so that its mass number reduces by 4 and atomic number reduces by 2 to become the unstable nucleus ^{A– 4}B* _{Z– 1}. It then emits a γ photon which does not produce any change in mass number and atomic number. The correct option is (4).

2. The number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it. The resulting daughter is an

(1) isomer of parent

(2) isotone of parent

(3) isotope of parent

(4) isobar of parent

When an α -particles emitted the mass number reduces by 4 and atomic number reduces by 2. When two β-particles are emitted the atomic number increases by 2. Therefore, when the number of β-particles emitted by a radioactive substance is equal to twice the number of α -particles emitted by it there is no change in the atomic number, but there is a reduction in mass number.

The resulting daughter is therefore an isotope of the parent nucleus [Option (3)].

3. In a Rutherford scattering experiment when a projectile of charge z₁ and mass M₁ approaches a target nucleus of charge z₂ and mass M₂,the distance of closest approach is r₀. The energy of the projectile is

(1) directly proportional to z₁z₂

(2) inversely proportional to z₁

(3) directly proportional to mass M₁

(4) directly proportional to M₁M₂

At the distance of closest approach the entire kinetic energy (E) of the projectile gets converted into electrostatic potential energy of the system. Therefore we have

E = (1/4πε₀) (z₁z₂/r₀)

Therefore, the energy of the projectile is directly proportional to z₁z₂ [Option (1)].