Given, T1/2 = 28 years = 28 × 3.154 × $10^7$ s Mass, m = 15 mg = 0.015 g Number of atoms in 0.015 g sample of $\,\mathrm{^{90}_{38}Sr}$ , N = $\frac{m}{M}$ × ×Avogadro's number = $\frac{0.015 \times 0.023 \times 10^{23} \text{ atoms}}{90}$ Activity of the sample, R = λN = $\frac{0.693}{T_{1/2}}$ N = $\frac{0.693 \times 0.015 \times 6.023 \times 10^{23}}{28 \times 3.154 \times 10^7 \times 90}$ = 7.877 × $10^{10}$ disintegration/sec = 7.877 × $10^{10}$ Bq = $\frac{7.877 \times 10^{10}}{3.7 \times 10^{10}}$ Ci = 2.13 Ci