A) 4 and 6
B) 6 and 4
C) 3 and 2
D) 2 and 3
Correct Answer: B
Solution :
\[_{90}T{{h}^{232}}{{\xrightarrow[{}]{{}}}_{82}}P{{b}^{208}}\] Number of a-particles\[=\frac{232-208}{4}=6\] Number of P-particles\[=6\times 2-(90-82)=4\]You need to login to perform this action.
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