Question

$10g$ atoms of an $\alpha$-active radioisotope are disintegrating in a sealed container. In one hour, the helium gas collected at STP is $11.2{cm}^3$. Calculate the half-life of the radioisotope.

Answer 1

Amount of radioactive isotope $=10g$ atoms

or $N_0=10\times 6.023\times {10}^{23}$ atoms

$=6.023\times {10}^{24}$ atoms

$22400{cm}^3$ of helium contains $=6.023\times {10}^{23}$ atoms
$11.2{cm}^3$ of helium will contain $=\frac{6.023\times {10}^{23}}{22400}\times 11.2$ atoms

$=3.01\times {10}^{20}$ atoms

As one helium atom is obtained by the disintegration of one atom of radioisotope, the total number of atoms of the radioactive isotope which have disintegrated in one hour.

$=3.01\times {10}^{20}$ or $\times 0.0003\times {10}^{24}$

The number of atoms of the radioactive isotope left after one hour,
$N=(6.023\times {10}^{24}-0.0003\times {10}^{24})$

$=6.0227\times {10}^{24}$

Using, $\lambda =\frac{2.303}{t}\log \frac{N_0}{N}$

$\lambda =\frac{2.303}{t}\log \frac{6.023\times {10}^{24}}{6.0227\times {10}^{24}}$

$=2.303\times 2.1632\times {10}^{-5}=4.982\times {10}^{-5}{hr}^{-1}$

$t_{1/2}=0.693/(4.982\times {10}^{-5}\times 24\times 365)=1.58$ years.