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  • 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.

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

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Required activity=8.0 mCi

1 Ci=3.7\times1010 decay s-1

8.0 mCi=8\times10-3\times3.7\times1010 =2.96\times108 decay s-1

T1/2=5.3 years

\lambda =\frac{0.693}{T_{1/2}}

\lambda =\frac{0.693}{5.3\times 365\times 24\times 3600}

\lambda =4.14\times 10^{-9}\ s^{-1}

\\\frac{\mathrm{d} N}{\mathrm{d} t}=-N\lambda \\ N=-\frac{\mathrm{d} N}{\mathrm{d} t}\times \frac{1}{\lambda }\\ N=-(-2.96\times 10^{8})\times \frac{1}{4.14\times 10^{-9}}\\ N=7.15\times 10^{16}\ atoms

Mass of those many atoms of Cu will be

w=\frac{7.15\times 10^{16}\times 60}{6.023\times 10^{23}}

w=7.12\times10^{-6} g

7.12\times10-6  g of _{27}^{60}\textrm{Co}  is necessary to provide a radioactive source of 8.0 mCi strength.

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