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 = 0.693/t_"1/2"`

`0.693/(3.82 xx 10^8) = 0.1814 xx 10^-8`

`= 1.81 xx 10^-9  "s"^-1`

No. of atoms left undecayed, N = n × 6.023 × 10²³

`= (5 xx 0.125)/(8.2 xx 10^-2 xx 3 xx 10^2) xx 6.023 xx 10^23...............(because n = "PV"/"RT")`

`= 1.5 xx 10^22` atoms

Activity, A = λN

`= 1.81 xx 10^-9 xx 1.5 xx 10^22 = 2.7 xx 10^13 " disintegration/sec"`

`therefore A = (2.7 xx 10^13)/(3.7 xx 10^10) "Ci" = 729.72` `"Ci"`