A) 4
B) 5
C) 6
D) 3
Correct Answer: B
Solution :
\[_{90}{{X}^{232}}\xrightarrow{-2\beta }{{\,}_{92}}{{Y}^{232}}\to {{\,}_{82}}{{Z}^{212}}+x{{\,}_{2}}H{{e}^{4}}\] No. of \[\alpha \]-particles \[=\frac{232-212}{4}=\frac{20}{4}=5\].You need to login to perform this action.
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