Question

${}_{84}P{o}^{210}$ (half-life= 138.4 days) undergoes $\alpha$- decay to form ${}_{82}P{b}^{206}$. If 2.1 g of ${}_{84}P{o}^{210}$ is placed in a sealed tube, what volume (in mL) of helium will accumulate in 69.2 days at STP?(take $\sqrt{2}$=1.4).

Answer 1

The radio active decay follows first order kinetics.
$t=\frac{1}{\lambda }ln\frac{N_0}{N_t}\Rightarrow 69.2=\frac{t_{1/2}}{ln2}ln\frac{N_0}{N_t}\Rightarrow 69.2=\frac{138.4}{ln2}\times ln\frac{N_0}{N_t}\Rightarrow ln2=ln{\left(\frac{N_0}{N_t}\right)}^2$

$\Rightarrow \frac{N_0}{N_t}=\sqrt{2}\Rightarrow N_t=\frac{\sqrt{2}}{2}{N}_0=\frac{1.4}{2}\times \frac{2.1}{210}=7\times {10}^{-3}$ moles of ${}_{84}P{o}^{210}$

So, moles of $Po$ decayed = $\frac{2.1}{210}-7\times {10}^{-3}=3\times {10}^{-3}$ mol

So, moles of He formed = $3\times {10}^{-3}$ mol

So, volume of He in ml=$3\times {10}^{-3}\times 22.4\times 1000ml=67.2ml$ at STP