In a radioactive decay chain, the initial nucleus is $_{90}^{232} \mathrm{Th}$. 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 :

(C) $\mathrm{A}=208 ; \mathrm{Z}=82$

(A) $\mathrm{A}=208 ; \mathrm{Z}=80$

(B) $\mathrm{A}=202 ; \mathrm{Z}=80$

(D) $\mathrm{A}=200 ; \mathrm{Z}=81$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Modern Physics

In a radioactive decay chain, the initial nucleus is

_{90}^{232} Th

. At the end there are

6 α

particles and

4 β

-particles which are emitted. If the end nucleus is

_{Z}^{A} X, A

and

Z

are given by :

(A) $A = 208; Z = 80$
(B) $A = 202; Z = 80$
(C) $A = 208; Z = 82$ ✓ Correct
(D) $A = 200; Z = 81$

Answer: (C)

Solution

Each

α

emission drops the mass number by

4

and the atomic number by

2

; each

β

emission raises the atomic number by

1

with the mass number unchanged. Starting from

_{90}^{232} Th

, six

α

give

A = 232 - 24 = 208

and

Z = 90 - 12 = 78

; the four

β

then raise

Z

78 + 4 = 82

with

A

still

208

. So

A = 208

Z = 82

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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