Ncert Exercises & Solutions

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

Hint: The rate of disintegration is given by: \(\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}\)

Step 1: Find the number of atoms of \({ }_{27}^{60} \mathrm{Co}\).
The strength of the radioactive source is given as:

\frac{dN}{dt} = 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

{}_{27}^{60}{Co}, T_{1 / 2} = 5.3 years = 5.3 \times 365 \times 24 \times 60 \times 60 = 1.67 \times 10^{8} \sec

For decay content

λ

, we have the rate of decay as:

\frac{dN}{dt} = λN

where,

λ = \frac{0 . 693}{T_{1 / 2}} = \frac{0.693}{1.67 \times 10^{8}} s^{- 1}

∴ N = \frac{1}{λ} \frac{dN}{dt}

= \frac{29.6 \times 10^{7}}{0.693} = 7.133 \times 10^{16}

atoms

Step 2: Find the amount of \({ }_{27}^{60} \mathrm{Co}\).
For

{}_{27}^{60}{Co}

, mass of

6.023 \times 10^{23}

(Avoadros number) atoms = 60 g

∴

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} g

Hence, the amount of

{}_{27}^{60}{Co}

necessary for the purpose is

7.106 \times 10^{- 6} g .

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