Half life of $_{38}^{90} S r, t_{1 / 2}=28$ years

$=28 \times 365 \times 24 \times 60 \times 60$

$=8.83 \times 10^{8} s$

Mass of the isotope, $m=15 mg$

$90 g$ of $_{38}^{90} Sr$ atom contains $6.023 \times 10^{23}(\text { Avogadro's number })$ atoms. Therefore, $15 mg$ of $_{38}^{90} Sr$ contains:

$\frac{6.023 \times 10^{2} \times 15 \times 10^{-3}}{90},$

i.e., $1.0038 \times 10^{20}$ Number of atomms

Rate of disintegration, $\frac{d N}{d t}=\lambda N$

Where, $\lambda=$ decay constant $=\frac{0.693}{8.83 \times 10^{8}} s^{-1}$

$\therefore \frac{d N}{d t}=\frac{0.693 \times 1.0038 \times 10^{20}}{8.83 \times 10^{8}}=7.878 \times 10^{10}$ atoms $/ s$

Hence, the disintegration rate of $15 mg$ of the given isotope is $7.878 \times 10^{10}\; atoms / s$