Question

Calculate the mass of $C^{14}$ (half life $=5720$ years) atoms which give $3.7\times {10}^7$ disintegrations per second.

Answer 1

Let the mass of ${}^{14}C$ atoms be $mg$.
Number of atoms in $mg$ of ${}^{14}C=\frac{m}{14}\times 6.02\times {10}^{23}$
$\lambda =\frac{0.693}{halflife}=\frac{0.693}{5720\times 365\times 24\times 60\times 60}=3.84\times {10}^{-12}{sec}^{-1}$
We know that, $-\frac{dN}{dt}=\lambda N$
i.e., Rate of disintegration $=\lambda \times$ number of atoms
$3.7\times {10}^7=\frac{0.693}{5720\times 365\times 24\times 60\times 60}\times \frac{m}{14}\times 6.02\times {10}^{23}$
$3.7\times {10}^7=\frac{3.84\times {10}^{-12}\times m\times 6.02\times {10}^{23}}{14}$
$m=2.24\times {10}^{-4}g$