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 α 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)

$N_{\beta }=2$

We are given that
${}_{90}^{232}Th\to {}_{82}^{212}Pb{+}_2^{4}He+{\beta }^{-}$
${}_{90}^{232}Th\to {}_{82}^{212}Pb{+}_2^{4}He{+}_{-1}{e}^{-}$
Let us conserve the mass on both the sides
$232=212+x\times 4$
$⟹x=5$
Thus, the number of alpha particles = 5
In order to find the number of beta particles, we also need to conserve the charge on both sides
$90=82+5\times 2+y(-1)$
$⟹y=2$
No. of beta particles = 2