The half-life of $_{38}^{90} Sr$ is $28$ years. What is the disintegration rate of $15\; mg$ of this isotope?
The half-life of $_{38}^{90} Sr$ is $28$ years. What is the disintegration rate of $15\; mg$ of this isotope?
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 $7.878 \times 10^{10}\; atoms / s$
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