Find the half-life of ${ }^{238}\mathrm{U} \left(\text{ in }\times {10}^9\right.$ years), if 1 g of it emits $1.24\times {10}^4\alpha -$particles per second. Avogadro's number = $6.023\times {10}^{23}$.

Here, $\mathrm{R}=1.24\times {10}^4\mathrm{dis}/\mathrm{s}$

Number of nuclei in 1 g of ${ }^{238}\mathrm{U},\text{ i.e., }\mathrm{N}=\frac{6.023\times {10}^{23}}{238}$

As $\mathrm{R}=\mathrm{\lambda N},$

$\mathrm{\lambda }=\frac{\mathrm{R}}{\mathrm{N}}=\frac{\left(1.24\times {10}^4\right)\times 238}{6.023\times {10}^{23}}=4.9\times {10}^{-18}{\mathrm{s}}^{-1}$

Thus, ${\mathrm{T}}_{1/2}=\frac{0.693}{\mathrm{\lambda }}=\frac{0.693}{4.9\times {10}^{-18}{\mathrm{s}}^{-1}}=1.41\times {10}^{17}\mathrm{s}$

$\Rightarrow \mathrm{T}=4.5\times {10}^9\text{ y (as 1 }\mathrm{y}=3.15\times {10}^7\mathrm{s}\text{ ) }$