Question

${}^{90}Sr$ shows $\beta$-activity and its half life period is $28$ years. What is the activity of a sample containing $1g$ of ${}^{90}Sr$?

Answer 1

Activity $=$ Number of atoms disintegrating per second

$=\lambda \times$ total number of atoms

$\lambda =\frac{0.693}{28\times 365\times 24\times 60\times 60}$

Total number of atoms in $1g$ of ${}^{90}Sr=\frac{6.023\times {10}^{23}}{90}$

Activity $=\frac{0.693}{28\times 365\times 24\times 60\times 60}\times \frac{6.023\times {10}^{23}}{90}$

$=5.25\times {10}^{12}$ disintegrations per second

$=\frac{5.25\times {10}^{12}}{3.7\times {10}^{10}}=141.89$ curie