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  4. The radioactive decay constant of 3890 S r is 7.88 × 10-10 s-1. The activity of 15 mg of this isotope will be

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Q. The radioactive decay constant of ${ }_{38}^{90} S r$ is $7.88 \times 10^{-10}\, s^{-1}$. The activity of $15 \,mg$ of this isotope will be

AMUAMU 2011

Solution:

Activity $\lambda . N$
$=7.88 \times 10^{-10} \times \frac{15 \times 10^{-3}}{90} \times 6.023 \times 10^{23} $
$=7.91 \times 10^{-10} d p s $
$=\frac{7.91 \times 10^{10}}{3.7 \times 10^{10}} $
$=2.13 \,Ci$