The half-life of

Half life of ${ }_{38}^{90} \mathrm{Sr}, t_{1 / 2}=28$ years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 108 s

Mass of the isotope, m = 15 mg

$90 \mathrm{~g}$ of ${ }_{38}^{90} \mathrm{Sr}$ atom contains $6.023 \times 10^{23}$ (Avogadro's number) atoms.

Therefore, $15 \mathrm{mg}$ of ${ }_{38}^{90} \mathrm{Sr}$ contains:

$\frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}{90}$, i.e., $1.0038 \times 10^{20}$ number of atoms

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

Where,

$\lambda=$ Decay constant $=\frac{0.693}{8.83 \times 10^{8}} \mathrm{~s}^{-1}$

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

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