The half life of $Sr_{38}^{90}$ is 28 years. The disintegrating rate of 15mg of this isotope is(Given 1 year =$ 3.15\times 10^7 s$:)
Answer & explanation
Correct answer: option 2
$ \text{Activity of the nucleus is } A = \lambda N$
$ \lambda = \frac{0.693}{t_\frac{1}{2}} = \frac{0.693}{28\times 3.15\times 10^7} s^{-1}$
$ N = \frac{15\times 10^{-3}}{90} \times 6.023 \times 10^{23} = 10^{20}$
$ A = \lambda N = \frac{0.693}{28\times 3.15\times 10^7}\times 10^{20} = 7.85\times 10^{10}Bq$