Question

1 g-atom of ${}^{226}\text{Ra}$ is placed in an evacuated tube of volume 5 litre. Assuming that each ${}_{88}^{226}\text{Ra}$ nucleus is an $\alpha$-emitter and all the contents are present in tube. The partial pressure of He collected in a tube at ${27}^{\circ }C$ after the end of 800 years is :

($t_{1/2}$ of Ra is 1600 year. Neglect volume occupied by undecayed Ra)

• A. $1.443atm$
• B. $2.443atm$
• C. $2.886atm$
• D. $0.721atm$

Answer 1

The correct option is A $1.443atm$

${}_{88}^{226}\text{Ra}{⟶}_{86}^{222}\text{Rn}{+}_2^4\text{He}$

$N_0=1g-atom;t_{1/2}Ra=1600year,t=800year$

Now, $t=\frac{2.303}{K}\log \frac{N_0}{N}$

$800=\frac{2.303\times 1600}{0.693}\log \frac{1}{N}(\therefore N=0.707g-atom)$

$\therefore$ Amount of Ra decayed $=1-0.707=0.293g-atom$
$\therefore$ Mole of Rn formed $=0.293andmoleofHeformed=0.293$
$\because PV=nRT$
$\therefore$ Total pressure of He and Rn is
$P=\frac{2\times 0.293}{5}\times 0.0821\times 300=2.887atm$
$\therefore P_{He}^{\prime }=P\times molefractionofHe$
$=2.887\times \frac{1}{2}=1.443atm$