# Obtain the amount of

Question:

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

Solution:

The strength of the radioactive source is given as:

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

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

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

Where,

N = Required number of atoms

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

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

For decay constant λ, we have the rate of decay as:

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

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

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

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

For ${ }_{27} \mathrm{Co}^{60}$ :

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

$\therefore$ Mass of $7.133 \times 10^{16}$ atoms $=\frac{60 \times 7.133 \times 10^{16}}{6.023 \times 10^{23}}=7.106 \times 10^{-6} \mathrm{~g}$

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