Question

The half-life of ${}_{38}^{90}Sr$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer 1

Given,

$m=15mg$

$T_{1/2}=28years=28\times 365\times 24\times 60\times 60=8.83\times {10}^{8}sec$

Half-life , $T_{1/2}=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{T_{1/2}}$

$\lambda =\frac{0.693}{8.83\times {10}^8}=7.84\times {10}^{-10}$

$15mg$ of ${}_{38}^{90}Sr$ contaion number of atoms

$N=\frac{15\times {10}^{-3}\times 6.022\times {10}^{23}}{90}=1.003\times {10}^{20}atom$

Rate of integration,

$\frac{dN}{dt}=\lambda N$

$\frac{dN}{dt}=7.84\times {10}^{-10}\times 1.003\times {10}^{20}$

$\frac{dN}{dt}=7.87\times {10}^{10}atoms/s$