Question

Obtain the amount of ${}_{27}^{60}Co$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}Co$ is 5.3 years.

Answer 1

Given,

$T_{1/2}=5.3year$

$T_{1/2}=5.3\times 365\times 24\times 60\times 60=1.67\times {10}^{8}sec$

$\frac{dN}{dt}=8.0mCI$

$\frac{dN}{dt}=8\times {10}^{-3}\times 3.7\times {10}^{10}=29.6\times {10}^{7}decay/sec$

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

$\lambda =\frac{0.693}{T_{1/2}}=\frac{0.693}{1.67\times {10}^8}$

$\lambda =0.414\times {10}^{-8}{s}^{-1}$

Radioactive rate, $\frac{dN}{dt}=\lambda N$

$N=\frac{1}{\lambda }.\frac{dN}{dt}$

$N=\frac{29.6\times {10}^7}{0.414\times {10}^{-8}}=7.15\times {10}^{16}atom$

For ${}_{27}^{60}Co$,

mass of $6.022\times {10}^{23}atom=60g$

mass of one atom $=\frac{60}{6.022\times {10}^{23}}g$

mass of one atom $7.15\times {10}^{16}atom\equiv \frac{60\times 7.15\times {10}^{16}}{6.022\times {10}^{23}}g=7.123\times {10}^{-6}g$

Hence, the amount of ${}_{27}^{60}Co$ is $7.123\times {10}^{-6}g$.