A) 4, 1
B) 3, 7
C) 8, 1
D) 4, 7
Correct Answer: A
Solution :
\[_{90}T{{h}^{228}}{{\to }_{83}}B{{i}^{212}}\] No. of \[\alpha \]-particles = \[\frac{228-212}{4}=\frac{16}{4}=4\] No. of \[\beta \]-particles = \[90-83-2\times 4=1\].You need to login to perform this action.
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