Question

In a radioactive decay chain, the initial nucleus is ${}_{90}{\text{Th}}^{232}$. At the end there are $6\alpha$ particles and $4{\beta }^{-}$ particles are emitted. If the end nucleus is ${}_Z{\text{X}}^A$, then $A\text{and}Z$ are given by :

• A. $A=208$; $Z=80$
• B. $A=202$; $Z=80$
• C. $A=200$; $Z=81$
• D. $A=208$; $Z=82$

Answer 1

The correct option is D $A=208$; $Z=82$

When $6\alpha$ particles are emitted, then,

${}_{90}{\text{Th}}^{232}\to {}_{90-(2\times 6)}{\text{Y}}^{232-(4\times 6)}+6{(}_2{\text{He}}^4)$

$\Rightarrow {}_{90}{\text{Th}}^{232}\to {}_{78}{\text{Y}}^{208}+6{(}_2{\text{He}}^4)$

Now, When $4$ more ${\beta }^{-}$ particles are emitted, then,

${}_{78}{\text{Th}}^{208}\to {}_{78+4\left(1\right)}{\text{X}}^{208}+4{(}_{-1}{e}^0)$

$\Rightarrow {}_{78}{\text{Th}}^{208}\to {}_{82}{\text{X}}^{208}+4{(}_{-1}{e}^0)$

Therefore, $A=208;Z=82$

Hence, option $\left(D\right)$ is correct.

Why this Question? To understand the $\alpha$ and $\beta$ decay with an example.