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Home » Obtain the amount of      necessary to provide a radioactive sourceof 8.0 mCi strength. The half-life of      is 5.3 years.
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.
CBSE | Class 12 | NCERT | Nuclei | Physics
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 –

We are given the strength of the radioactive source as –

    \[\frac{dN}{dt}=8.0mCi\]

= 8 x 10-3 x 3.7 x 1010

= 29.6 x 107 decay/s

Where,

N is the required number of atoms

Half-life of 

    \[{}_{27}^{60}Co\]

is given to us, which is = 5.3 years

Converting the half life time into seconds, we get –

    \[{{T}_{\frac{1}{2}}}\]

= 5.3 × 365 × 24 × 60 × 60

    \[{{T}_{\frac{1}{2}}}\]

= 1.67 × 108 s

We know that the rate of decay for decay constant λ can be written as following –

    \[\frac{dN}{dt}=\lambda N\]

    \[\Rightarrow \lambda =\frac{0.693}{{{T}_{\frac{1}{2}}}}\]

Using both relations, we get –

    \[N=\frac{1}{\lambda }\frac{dN}{dt}=\frac{1.67\times {{10}^{8}}\times 29.6\times {{10}^{7}}}{0.693}\]

N = 7.133 x 1016 atoms

For 

    \[{}_{27}^{60}Co\]

–

Mass of 6.023 × 1023 atoms = 60 g

Mass of 7.133 x 1016 atoms will be given by

    \[=\frac{60\times 7.133\times {{10}^{16}}}{6.023\times {{10}^{23}}}\]

= 7.106 x 10-6 g

Hence, the amount of 

    \[{}_{27}^{60}Co\]

necessary for the purpose is 7.106 × 10−6 g.

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