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

The strength of the radioactive source is given as

$\frac{d N}{d t}=8.0 mCi$

$=8 \times 10^{-3} \times 3.7 \times 10^{10}$

$=29.6 \times 10^{7}$ decay $/ s$

Where, $N =$ Required number of atoms

Half-life of $\frac{60}{27} Co , T_{1 / 2}=5.3$ years

$=5.3 \times 365 \times 24 \times 60 \times 60$

$=1.67 \times 108 s$

For decay constant $\lambda,$ we have the rate of decay as $\frac{d N}{d t}=\lambda N$

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

$\therefore N=\frac{1}{\lambda} \frac{d N}{d t}$

$=\frac{29.6 \times 10^{7}}{\frac{0.693}{1.67 \times 10^{8}}}=7.133 \times 10^{16}$ atoms

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

Mass of $6.023 \times 1023$ (Avogadro's number) atoms $=60 g$

Mass of $7.133 \times 10^{16}$ atoms $=\frac{60 \times 7.133 \times 10^{16}}{6.023 \times 10^{33}}=7.106 \times 10^{-6} g$

Hence, the amount of $_{27} Co ^{60}$ necessary for the purpose is $7.106 \times 10^{-6}\; g$