Answer:
Given mass number of lead\[={{A}_{Pb}}=208;\] atomic number of lead\[={{Z}_{Pb}}=82\] \[\therefore \]No. of neutrons in lead\[={{n}_{Pb}}={{A}_{Pb}}{{Z}_{Pb}}=\] \[20882=126\] \[\Rightarrow \]\[{{(n/P)}_{Bi}}=\frac{126}{83}=1.51\] _______ (1) Mass number of bismuth\[={{A}_{Bi}}=209\] Atomic number of bismuth\[={{Z}_{Bi}}=83\] \[\therefore \]No. of neutrons in bismuth \[={{n}_{Bi}}={{A}_{bi}}{{Z}_{Bi}}\] \[=20983=126\] \[\Rightarrow \]\[{{(n/P)}_{Bi}}=\frac{126}{83}=1.51\] ________ (2) From (a) and (b) the n/p ratio of lead is greater.
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