Solveeit Logo

Question

Physics Question on Nuclei

In a radioactive decay chain, the initial nucleus is 90232Th^{232}_{90}Th . At the end there are 6α6 \alpha-particles and 4β4 \beta-particles which are emitted. If the end nucleus, If ZAX^{A}_{Z} X , AA and ZZ are given by :

A

A = 208; Z = 80

B

A = 202; Z = 80

C

A = 200; Z = 81

D

A = 208; Z = 82

Answer

A = 208; Z = 82

Explanation

Solution

The correct answer is (D) : A = 208; Z = 82
When one α−particle is emitted, then the mass number (A) of daughter nuclei decreases by 4 and the atomic number decreases by 2.
^{232}_{90}Th$$→^{208}_{78}Y+6(^4_2He)
When one β−particle is emitted, then the mass number (A) of daughter nuclei increases by 1 and the atomic number remains the same.
78208Y82208X+4β^{208}_{78}Y→ ^{208}_{82}X+4β
Therefore, for the end nucleus, A = 208 : Z = 82