Modern Physics 3 Question 37

1. 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)

Show Answer

Solution:

  1. 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 $A=208, Z=82$



NCERT Chapter Video Solution

Dual Pane