Obtains the amount of (27Co60) necessary to provide a radioactive source of 8.0 m Ci strength. The half – life of (27Co60) is 5.3 years.

Q: Obtains the amount of (₂₇Co⁶⁰) necessary to provide a radioactive source of 8.0 m Ci strength. The half –
life of (₂₇Co⁶⁰) is 5.3 years.

Sol: Half –life of ₂₇Co⁶⁰ = 5.3 years

= 5.3 x 365 x 24 x 60 x 60 s = 5.3 x 3.15 x 10⁷ s

Now ‘ X ’ gm of ₂₇Co⁶⁰ contains 10^-3/60 K – mole

= (X × 10^-3)/60 × 6.025 × 10²⁶

N = 0.1004 X × 10²³ = 1.004 X × 10²² atoms

Required strength of ₂₇Co⁶⁰ = 8 m Ci

= 8 x 10^-3 × 3.7 × 10¹⁰ dis/s

We known that, decay rate (R) = λN

N = R/λ = (8 × 3.7 × 10⁷)/(0.693/(T_1/2)

= (29.6 × 10⁷)/0.693 × 5.3 × 3.15 × 10⁷

= 713.0909 × 10¹⁴

1.004 X × 10²² = 713.0909 × 10¹⁴

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