Question
Physics Question on Nuclei
In a radioactive decay chain, the initial nucleus is 90232Th . At the end there are 6α-particles and 4β-particles which are emitted. If the end nucleus, If ZAX , A and Z 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.
78208Y→82208X+4β
Therefore, for the end nucleus, A = 208 : Z = 82