Question

An atomic nucleus ${}_{90}T{h}^{232}$ emits several $\alpha$ and $\beta$ radiations and finally reduces to ${}_{82}P{b}^{208}$. It must have emitted.

• A. $4\alpha$ and $2\beta$
• B. $6\alpha$ and $4\beta$
• C. $8\alpha$ and $24\beta$
• D. $4\alpha$ and $16\beta$

Answer 1

The correct option is B $6\alpha$ and $4\beta$

$6\alpha$ and $4\beta$

Let $a$ and $b$ be the number of $\alpha$ and $\beta$ particles emitted during the reaction

${}_{90}T{h}^{232}{\to }_{82}P{b}^{208}+a$ ${}_{2}H{e}^4+b$ ${}_{-1}{e}^0$

Comparing the mass numbers,

$232=208+4a+(b$ X $0)$

$4a=232-208$

$a=\frac{24}{4}$

$a=6$

Comparing the atomic numbers

$90=82+(2$ X $a)+-1b$

$90=82+2a-b$

$2a-b=90-82=8$

Substituting the value of $a$ in the above equation we get,

$2$ X $46-b=8$

$b=12-8=4$

So,

The number of $\alpha$ particles emitted$=6$

The number of $\beta$ particles emitted $=4$