A) 4, 2
B) 2, 2
C) 8, 6
D) 6, 4
Correct Answer: D
Solution :
No. of \[\alpha \]-particles \[=\frac{1}{4}\times \] (Difference in mass number) \[=\frac{1}{4}\times (232-208)=\frac{1}{4}\times 24=6\] No. of \[\beta \]-particles \[=2\times \]\[\alpha \]-particles (Difference in atomic number) \[=2\times 6-(90-82)=12-8=14\]You need to login to perform this action.
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