Question

10 g of atoms of an $\alpha$ active radioactive isotope are disintegrating in a sealed container. In one hour, the He gas collected at STP is 11.2 $c{m}^3$. The half life of radioactive isotope(in hrs) is :

• A. 27820
• B. 6945
• C. 13910
• D. none of these
  

Answer 1

Answer: (C)

13910

$N_0=10gatoms=10\times 6.023\times {10}^{23}=6.023\times {10}^{24}atoms$

Volume of He collected $=11.2mL$.

$=\frac{11.2}{22400}mole$
$=5\times {10}^{-4}mole$
$=5\times {10}^{-4}\times 6.023\times {10}^{23}atoms$
$=3.01\times {10}^{20}atoms$

He atoms formed = Number of atoms of radioactive substance decayed. Number of atoms of radioactive substance left = N

$=6.023\times {10}^{24}-3.01\times {10}^{20}$
$=6.0227\times {10}^{24}atoms$

$K=\frac{2.303}{t}log\frac{6.023\times {10}^{24}}{6.0227\times {10}^{24}}=4.982\times {10}^{-5}h{r}^{-1}$

$t_{1/2}=\frac{0.693}{K}=\frac{0.693}{4.982\times {10}^{-5}}=13910.29hr$.... (i)

Note : $N_0$ and N can be put directly in terms of mole or g atoms but in this problem it will lead problem in solving log values).