Radioactivity — Question 8

(a) Due to emission of a β particle, the atomic number increases by 1 and the mass number is unchanged. Hence, the nuclear change is —

$_{x}^{a}\text{P} \longrightarrow _{x + 1}^{a}\text{Q} + _{-1}^{0}\text{β}$

(b) Due to α emission, atomic number decreases by 2 and mass number decreases by 4. Also, an alpha particle is same as a Helium nucleus with 2 protons and 2 neutrons. Hence, we get the nuclear change as —

$_{92}^{238}\text{U} \longrightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He} + \text{energy}$

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

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

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

So the complete nuclear change is as follows —

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

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

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

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 $_{Z}^{A - 4}\text{X}_{3}$

Hence, the nuclear change is as follows —

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

(e) The daughter nucleus $_{\space \space 69}^{172}\text{X}_{3}$ is formed after α decay of $\text{X}_{2}$, so atomic number of $\text{X}_{2}$ will be 2 units more and it's mass number will be 4 units more i.e., $_{\space \space 71}^{176}\text{X}_{2}$

The nucleus $_{\space \space 71}^{176}\text{X}_{2}$ is formed after α decay of $\text{X}_{1}$ so atomic number of $\text{X}_{1}$ will be 2 units more and it's mass number will be 4 units more i.e., $_{\space \space 73}^{180}\text{X}_{2}$

The nucleus $_{\space \space 73}^{180}\text{X}_{2}$ is formed after β decay of X, the atomic number will be one unit less and the mass number is unchanged i.e., $_{\space \space 72}^{180}\text{X}$

Hence, the nuclear change is as follows —

$_{\space \space 72}^{180}\text{X} \overset{\beta} \longrightarrow \space _{\space \space 73}^{180}\text{X}_{1} \overset{\alpha} \longrightarrow \space _{\space \space 71}^{176}\text{X}_{2} \overset{\alpha} \longrightarrow \space _{\space \space 69}^{172}\text{X}_{3}$