A) 5 and 4
B) 2 and 3
C) 3 and 2
D) 4 and 1
Correct Answer: C
Solution :
\[_{90}{{E}^{232}}\xrightarrow{{}}{{\,}_{86}}{{G}^{220}}\] No. of \[\alpha \] particle =\[\frac{232-220}{4}=3\] No. of \[\beta \] particle \[=86-[90-2\times 3]=2\]You need to login to perform this action.
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