Data : Activity = 10.0 mCi = 10.0 × 10-3 Ci
= (10.0 × 10-3 )(3.7 × 1010) dis/s = 3.7 × 108 dis/s
T1/2 = 5.3 years = (5.3)(3.156 × 107 ) s
= 1.673 × 108 s
Decay constant, \(\lambda\) = \(\cfrac{0.693}{T_{1/2}}\) = \(\cfrac{0.693}{1.673\times10^8}s-1\)
=4.142 × 10-9 s-1
Activity = Nλ
∴ N = \(\cfrac{activity}{\lambda}\) = \(\cfrac{3.7\times10^8}{4.142\times10^{-9}}\) atoms
= 8.933 × 1016 atoms
= 60 grams of \(^{60}_{27}Co\) contain 6.02 × 1023 atoms
Mass of 8.933 × 1016 atoms of \(^{60}_{27}Co\)
= \(\cfrac{8.933\times10^{16}}{6.02\times10^{23}}\) x 60 g
= 8.903 × 10-6 g = 8.903 µg
This is the required amount.