A) Higher of lead
B) Higher of bismuth
C) Same
D) None of these
Correct Answer: A
Solution :
\[\frac{n}{p}\] of \[_{82}P{{b}^{208}}=\frac{126}{82}=1.53\] \[\frac{n}{p}\] of \[_{83}B{{i}^{209}}=\frac{126}{83}=1.51\]You need to login to perform this action.
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