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Question
A vessel of volume 125 cm3 contains tritium (3H, t1/2 = 12.3 y) at 500 kPa and 300 K. Calculate the activity of the gas.
Solution
Given:-
Volume of the vessel, V = 125 cm3 = 0.125 L
Half-life time of tritium, t1/2 = 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"`
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