From helium-3 volume to accumulation time

Problem

A device contains helium-3 gas produced by the $\beta^-$ decay of tritium. Starting from $1\ \text{g}$ of tritium ($T_{1/2} = 12{,}3\ \text{years}$), a volume $V = 2{,}24\ \text{ml}$ of $^{3}\mathrm{He}$ is measured at STP. Knowing that the molar volume at STP is $22{,}4\ \text{L}$, how many years $N$ of accumulation have elapsed?

Solution

Helium-3 atoms produced. From the measured volume we obtain the moles of $^{3}\mathrm{He}$:

$n_{\text{He}} = \frac{V}{V_{\text{molar}}} = \frac{2{,}24\cdot 10^{-3}\ \text{L}}{22{,}4\ \text{L}} = 1{,}0\cdot 10^{-4}\ \text{mol}$

$N_{\text{He}} = n_{\text{He}}\,N_A = 1{,}0\cdot 10^{-4}\cdot 6{,}022\cdot 10^{23} = 6{,}02\cdot 10^{19}\ \text{atoms}$

Initial tritium atoms. The molar mass of tritium is $\approx 3{,}016\ \text{g/mol}$:

$N_0 = \frac{1}{3{,}016}\cdot 6{,}022\cdot 10^{23} \approx 2{,}0\cdot 10^{23}\ \text{atoms}$

Accumulation law. The helium-3 produced equals the tritium that has decayed: $N_{\text{He}} = N_0\left(1 - e^{-\lambda t}\right)$. The decayed fraction is:

$\frac{N_{\text{He}}}{N_0} = \frac{6{,}02\cdot 10^{19}}{2{,}0\cdot 10^{23}} \approx 3{,}0\cdot 10^{-4}$

Since the fraction is very small, we can linearise $1 - e^{-\lambda t} \approx \lambda t$:

$\lambda t \approx 3{,}0\cdot 10^{-4}, \qquad \lambda = \frac{\ln 2}{12{,}3} = 0{,}0564\ \text{years}^{-1}$

$t = \frac{3{,}0\cdot 10^{-4}}{0{,}0564} \approx 5{,}3\cdot 10^{-3}\ \text{years}$

$\ev{N \approx 0{,}0053\ \text{years} \approx 2\ \text{days}}$

Collegamenti

Argomenti: Fisica nucleare Concetti: Decadimento radioattivo Metodi: Separazione delle variabili