JEE PYQ: Nuclear Physics Question 28
Question 28 - 2019 (12 Jan Shift 1)
In a radioactive decay chain, the initial nucleus is $^{232}_{90}\text{Th}$. At the end there are 6 $\alpha$-particles and 4 $\beta$-particles which are emitted. If the end nucleus is $^A_Z\text{X}$, A and Z are given by:
(1) $A = 208; Z = 80$
(2) $A = 202; Z = 80$
(3) $A = 208; Z = 82$
(4) $A = 200; Z = 81$
Show Answer
Answer: (3)
Solution
When one $\alpha$-particle emitted then daughter nuclei has 4 unit less mass number (A) and 2 unit less atomic (Z) number ($z$). $^{232}{90}\text{Th} \to ^{208}{78}\text{Y} + 6 ^4_2\text{He}$ $^{208}{78}\text{Y} \to ^{208}{82}\text{X} + 4\beta$ particle
BETA
