Question

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

Answer 1

$-\frac{dN}{dt}=8mCi=8\times 3.7\times {10}^7=2.96\times {10}^8{s}^{-1}$

$T=5.3years=5.3\times 365\times 24\times 60\times 60=1.67\times {10}^{8}s$

$\therefore \lambda =\frac{0.693}{T}=\frac{0.693}{1.67\times {10}^8}=4.15\times {10}^{-9}{s}^{-1}$

$\frac{dN}{dt}=\lambda N\Rightarrow N=\frac{dN/dt}{\lambda }=\frac{2.96\times {10}^8}{4.15\times {10}^{-98}}=7.133\times {10}^{16}$

Mass of ${}_{27}^{60}Co,m=\frac{no.of{}_{27}^{60}Coatoms}{Avogadronumber}\times atomicweightof{}_{27}^{60}Co$

$m=\frac{7.133\times {10}^{16}}{6.02\times {10}^{23}}\times 60=7.11\times {10}^{-6}g$