STANDARDChemistry-Nuclear Chemistry
STANDARD Chemistry Radioactive Decay MCQ Question
Type: MCQ-conceptual-Medium
When ₉₀Th²³⁸ changes into ₈₃Bi²²², then the number of emitted α and β-particles are :-
A
8α, 7β
B
4α, 7β
C
4α, 4β
D
4α, 1β
Correct Answer
Option D
Detailed Explanation
The number of α-particles emitted nₐ = (A - A')/4 = (228 - 212)/4 = 4. The number of β-particles emitted nβ = 2nₐ - Z + Z' = 2 × 4 - 90 + 83 = 1.
Found an issue with this question?