Question

A sample of ${}_{38}^{90}Sr$ has an activity of $0.5mCi$. What is its specific activity? ($t_{1/2}$ of ${}_{38}^{90}Sr=19.9$ years)

Answer 1

Rate of disintegrations $=\lambda \times$ Number of atoms
So, Number of atoms $=\frac{0.5\times 3.7\times {10}^7}{0.693}\times 19.9\times 365\times 24\times 60\times 60$$=1.675\times {10}^{16}$
Mass $=\frac{90\times 1.675\times {10}^{16}}{6.023\times {10}^{23}}=2.50\times {10}^{-6}g$
Specific activity $=\frac{0.5\times 3.7\times {10}^7}{2.5\times {10}^{-6}}=7.4\times {10}^{12}dis{g}^{-1}{s}^{-1}$