Question

In a radioactive decay chain, ${}_{90}^{232}Th$ nucleus decays to ${}_{82}^{212}Pb$ nucleus. Let $N_{\alpha }$ and $N_{\beta }$ be the number of $\alpha$ and ${\beta }^{-}$ particles respectively, emitted in this decay process. Which of the following statements is (are) true?

• A. $N_{\alpha }=5$
• B. $N_{\alpha }=6$
• C. $N_{\beta }=2$
• D. $N_{\beta }=4$

Answer 1

Answer: (A), (C)

$N_{\beta }=2$

As given,
${}_{90}T{h}^{232}{\to }_{82}P{b}^{212}+X\alpha +Y\beta$ ....(x)

By charge conservation,
Total nuclear charge on $\alpha$ particle= 2 unit positive charge.
Total nuclear charge on $\beta$ particle= 1 unit negative charge.

taking the RHS part of equation (x)
$2X-Y+82=90$ .....(1)

On comparing the total mass of system,
${}_{90}T{h}^{232}{\to }_{82}P{b}^{212}+X_2{\alpha }^{4++}+Y_{-1}{\beta }^0$

then,
$4X=20$ .....(2)

By comparing (1) & (2)
$X=5$
and $2X-Y+82=902\times 5-Y=8Y=2$

Hence, $N_{\alpha }=5$ and $N_{\beta =2}$
so, option A and C is correct.