Saturday, April 11, 2009

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:

AXZ AYZ+1A4 B* Z1 A4 B Z1,

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 AXZ emits a β-particle and its atomic number increases by 1 to transform to the nucleus AYZ+1. The nucleus AYZ+1 emits an α-particle so that its mass number reduces by 4 and atomic number reduces by 2 to become the unstable nucleus A4B* Z1. 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)].