Answer to Question #121465 in Atomic and Nuclear Physics for Dheeraj

Question

The rate of electron emission from 4mg of Pb (atomic numer=80, atomic mass=210) wwith half life of 5 days is?

Answer 1

moles of Pb here $\frac{4\times10^{-3}}{210}=1.9\times10^{-5}$

moles of electron here = $80\times1.9\times10^{-5}$ =$1.52\times10^{-3}$

$k=\frac{\ln2}{t_{1/2}}=\frac{0.69}{5}=1.38day^{-1}$

rate of electron emission = $kA=1.38\times1.52\times10^{-3}=2.1\times10^{-3}eletrons/day$