Question

A vessel of volume 125 cm³ contains tritium (³H, t_1/2 = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.

Answer 1

Given:
Volume of the vessel, V = 125 cm³ = 0.125 L
Half-life time of tritium, t_1/2 = 12.3 y = 3.82 × 10⁸ s
Pressure, P = 500 kpa = 5 atm
Temperature, T = 300 K,

Disintegration constant, $\lambda =\frac{0.693}{t_{1/2}}$
=$$\begin{gathered} \frac{0.693}{3.82\times {10}^8}=0.1814\times {10}^{-8} \\ =1.81\times {10}^{-9}{\mathrm{s}}^{-1} \end{gathered}$$

No. of atoms left undecayed, N = n × 6.023 × 10²³
$$\begin{gathered} =\frac{5\times 0.125}{8.2\times {10}^{-2}\times 3\times {10}^2}\times 6.023\times {10}^{23}\left(\because n=\frac{PV}{RT}\right) \\ =1.5\times {10}^{22}\mathrm{atoms} \end{gathered}$$

Activity, A = λN
= 1.81 × 10⁻⁹ × 1.5 × 10²² = 2.7 × 10¹³ disintegration/sec

$\therefore A=\frac{2.7\times {10}^{13}}{3.7\times {10}^{10}}\mathrm{Ci}=729.72\mathrm{Ci}$