Solved Sample Paper 5 — Question 16

(a) If the moment of force is assigned a negative sign, then the turning tendency of the force will be in clockwise direction.

(b)

I. When there is an α emission from $^{238}_{92}\text{P}$ , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get $^{234}_{90}\text{Q}$ .

Now, when $^{234}_{90}\text{Q}$ undergoes β emission, the atomic number increases by 1 and the mass number is unchanged. Hence, we get $^{234}_{91}\text{R}$.

In the final step when $^{234}_{91}\text{R}$ undergoes β emission again the atomic number increases by 1 and the mass number is unchanged. Hence we get $^{234}_{92}\text{S}$.

So the complete nuclear change is as follows —

$^{238}_{92}\text{P} \overset{\alpha} \longrightarrow \space ^{234}_{90}\text{Q} \overset{\beta} \longrightarrow \space ^{234}_{91}\text{R} \overset{\beta} \longrightarrow \space ^{234}_{92}\text{S}$

II. When there is an α emission from $^{A}_{Z}\text{X}$ , then the atomic number decreases by 2 and mass number decreases by 4. Hence, we get $^{A - 4}_{Z - 2}\text{X}_{1}$

After the γ emission, there is no change in atomic number and mass number. Hence, we get $^{A - 4}_{Z - 2}\text{X}_{2}$

In the final step when there are two β emissions, the atomic number increases by 2 and the mass number is unchanged and we get $^{A - 4}_{Z}\text{X}_{3}$

Hence, the nuclear change is as follows —

$^{A}_{Z}\text{X} \overset{\alpha} \longrightarrow ^{A - 4}_{Z - 2}\text{X}_{1} \overset{\gamma} \longrightarrow ^{A - 4}_{Z - 2}\text{X}_{2} \overset{2\beta} \longrightarrow _{\space \space \space \space \space Z}^{A - 4}\text{X}_{3}$