Complete the following nuclear changes —

(a) $_{x}^{a}\text{P} \longrightarrow Q + _{-1}^{0}\text{β}$
n
(b) $_{92}^{238}\text{U} \longrightarrow \space _{90}^{234}\text{Th} + …… + \text{energy}$

(c) $_{92}^{238}\text{P} \overset{\alpha} \longrightarrow \text{Q} \overset{\beta} \longrightarrow \text{R} \overset{\beta} \longrightarrow \text{S}$

(d) {Z}^{A}\text{X} \overset{\alpha} \longrightarrow \text{X}{1} \overset{\gamma} \longrightarrow \text{X}{2} \overset{2\beta} \longrightarrow \text{X}{3}

(e) $\text{X} \overset{\beta} \longrightarrow \text{X}${1} \overset{\alpha} \longrightarrow \text{X}{2} \overset{\alpha} \longrightarrow \space {69}^{172}\text{X}{3}

(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 ${}_{92}^{238}\text{P}${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 ${}_{90}^{234}\text{Q}${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 ${}_{91}^{234}\text{R}${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 ${}_Z^A\text{X}${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 ${}_{Z-2}^{A-4}{\text{X}}_2${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 ${}_Z^{A-4}{\text{X}}_3${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 ${}_{69}^{172}{\text{X}}_3${\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., ${}_{71}^{176}{\text{X}}_2${\space \space 71}^{176}\text{X}{2}

The nucleus ${}_{71}^{176}{\text{X}}_2${\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., ${}_{73}^{180}{\text{X}}_2${\space \space 73}^{180}\text{X}{2}

The nucleus ${}_{73}^{180}{\text{X}}_2${\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 —

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