Q: Obtains the amount of (_{27}Co^{60}) necessary to provide a radioactive source of 8.0 m Ci strength. The half –

life of (_{27}Co^{60}) is 5.3 years.

Sol: Half –life of _{27}Co^{60} = 5.3 years

= 5.3 x 365 x 24 x 60 x 60 s = 5.3 x 3.15 x 10^{7} s

Now ‘ X ’ gm of _{27}Co^{60} contains 10^{-3}/60 K – mole

= (X × 10^{-3})/60 × 6.025 × 10^{26}

N = 0.1004 X × 10^{23} = 1.004 X × 10^{22} atoms

Required strength of _{27}Co^{60} = 8 m Ci

= 8 x 10^{-3} × 3.7 × 10^{10} dis/s

We known that, decay rate (R) = λN

N = R/λ = (8 × 3.7 × 10^{7})/(0.693/(T_{1/2})

= (29.6 × 10^{7})/0.693 × 5.3 × 3.15 × 10^{7}

= 713.0909 × 10^{14}

1.004 X × 10^{22} = 713.0909 × 10^{14}

X = 710. 2499 × 10^{-8} Kg = 7.1 × 10^{-6} g