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:

^{A}X_{Z }→_{ }^{A}Y_{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^{ A}X_{Z} emits a β-particle and its atomic number increases by 1 to transform to the nucleus ^{A}Y_{Z+1}. The nucleus ^{A}Y_{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 _{1} and mass M_{1} approaches a target nucleus of charge z_{2} and mass M_{2},the distance of closest approach is r_{0}. The energy of the projectile is

(1) directly proportional to z_{1}z_{2}

(2) inversely proportional to z_{1}

(3) directly proportional to mass M_{1}

(4) directly proportional to M_{1}M_{2}

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 = (1/4πε_{0}) (z_{1}z_{2}/r_{0})

Therefore, the energy of the projectile is directly proportional to z_{1}z_{2} [Option (1)].

## No comments:

## Post a Comment