Question

The number of $\alpha$ and $\beta$ particles emitted in the conversion of ${}_{90}T{h}^{232}$ to ${}_{82}P{b}^{208}$ are

• A. $6,4$
• B. $4,6$
• C. $8,6$
• D. $6,8$

Answer 1

The correct option is A $6,4$

(i) During $\alpha -$ emission:

${}_{90}{T}^{232}{\to }_{82}P{b}^{208}+{(}_{2}H{e}^4)x+y{(}_{-1}{e}^0)$

Equation of mass number,

$232=208+4x$

$\Rightarrow 232-208=4x$

$\therefore x=\frac{24}{4}=6$

(ii) During ${\beta }^{-}$ emission, mass number will not change, but atomic number will increase by $1$,

Equating atomic number,

$90-6\left(2\right)+y\left(1\right)=82$

$\Rightarrow y=82-78=4$

Hence, $\left(A\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this Question?}\\ \text{This type of problems are frequently asked in JEE Mains.}\end{array}$$