Modern Physics Ques 136
- In a radioactive decay chain, the initial nucleus is ${ } _{90}^{232} Th$. At the end, there are $6 \alpha$-particles and $4 \beta$-particles which are emitted. If the end nucleus is ${ } _Z^{A} X, A$ and $Z$ are given by
(a) $A=202 ; Z=80$
(b) $A=208 ; Z=82$
(c) $A=200 ; Z=81$
(d) $A=208 ; Z=80$
(Main 2019, 12 Jan II)
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Solution:
Formula:
- An $\alpha$-particle decay $\left({ } _2^{4} He\right)$ reduces, mass number by 4 and atomic number by 2 .
$\therefore$ Decay of $6 \alpha$-particles results
$$ { } _{90}^{232} Th \quad \stackrel{6 \alpha}{\rightarrow}{ } _{90-12}^{232-24} Y={ } _{78}^{208} Y $$
A $\beta$-decay does not produces any change in mass number but it increases atomic number by 1 .
$\therefore$ Decay of $4 \beta$-particles results
$$ { } _{78}^{208} Y \quad \stackrel{4 \beta}{\rightarrow}{ } _{82}^{208} X $$
$\therefore$ In the end nucleus $A=208, Z=82$
BETA
