Question

How many mL of radon (Rn) under a standard condition of temperature and pressure are in equilibrium with 1 g of radium (Ra)?

[Given half-life period of ${}^{226}Ra=1590yr$ and for ${}^{222}Rn=3.82$ days.]

• A. $6.52\times {10}^{-4}$ mL
• B. $8.52\times {10}^{-4}$ mL
• C. $7.52\times {10}^{-4}$ mL
• D. None of these

Answer 1

The correct option is A $6.52\times {10}^{-4}$ mL

At equilibrium,

$\frac{N_1\left(Rn\right)}{N_2\left(Ra\right)}=\frac{{\lambda }_2\left(Ra\right)}{{\lambda }_1\left(Rn\right)}=\frac{t_{\frac{1}{2}}\left(Rn\right)}{t_{\frac{1}{2}}\left(Ra\right)}$

$\frac{N_1\left(Rn\right)}{\left(1/226\right)mol}=\frac{3.82days}{365\times 1590days}$

$N_1\left(Rn\right)=2.91\times {10}^{-8}$ mol

1 mole of a gas at STP $=22400ml$

$\therefore$ Volume of $Rn$ at STP $=6.52\times {10}^{-4}$ mL

Hence, option A is correct.