Question

10 g-atoms of an $\alpha$-particle radioactive isotope are disintegrating in a sealed container. In one hour, the He gas collected at STP is $11.2c{m}^3$.

Half life of radio isotope in hours is:

Answer 1

Sol I : Given, Initial amount of radioactive species $N_0=10g-atom=10\times 6.023\times {10}^{23}atoms$
$=6.023\times {10}^{24}atoms$

No. of He-atoms collected $=\frac{\text{Volume of He-atoms collected}\times \text{Avogadro No.}}{22400}$
$=\frac{11.2\times 6.023\times {10}^{23}}{22400}=3.01\times {10}^{20}$ atoms

Now, No. of atoms of radioactive species decayed
$=$No. of He atoms formed$=3.01\times {10}^{20}$ atoms.

No. of radioactive species left$=N_0-$No. of atoms decayed $=6.023\times {10}^{24}-3.01\times {10}^{20}=6.0227\times {10}^{24}$

The decay constant $\lambda =\frac{2.303}{t}\log \frac{N_0}{N}=\frac{2.303}{1}\log \frac{6.023\times {10}^{24}}{6.0227\times {10}^{24}}$
$=4.982\times {10}^{-5}h{r}^{-1}$

The half life period $t_{1/2}=\frac{0.693}{\lambda }=\frac{0.693}{4.982\times {10}^{-5}}=13910.29hr$