Question

A vessel of volume 125 $c{m}^3$ contains tritium ${(}^{3}H,t_{\frac{1}{2}}$ = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.

Answer 1

Given V = 125 $c{m}^3$ = 0.125 L,

P = 500 Kpa = 5 atm

T = 300 K,

$t_{\frac{1}{2}}$ = 12.3 year

= $3.82\times {10}^8$ sec.

$\therefore$ N = n $\times 6.023\times {10}^{2}3$

= $\frac{5\times 0.125}{8.2\times {10}^{-2}\times 3\times {10}^2}\times 6.023\times {10}^{23}$

= $1.5\times {10}^{22}$ atoms

$\lambda =\frac{0.693}{3.82\times {10}^6}$

= $0.1814\times {10}^{-8}=1.81\times {10}^{-9}{s}^{-1}$

$\therefore$ Activity = $\lambda$ N

= $1.81\times {10}^{-9}\times 1.5\times {10}^{2}2\times {10}^{13}$

$\frac{\text{disintegration}}{\text{sec}}$.

= $\frac{2.7\times {10}^{13}}{3.7\times {10}^{10}}$ Ci = 727 Ci