1.

The half life period of ._(58)^(141)Ce is 13.11 days. It is a beta-particle emitter and the average energy of the beta-particle emitted is 0.442 MeV. What is the total energy emitted per second in watts by 10 mg of ._(58)^(141)Ce?

Answer»

SOLUTION :Rate of disintegration per sec =` LAMBDA xx` No. atoms
`= (0.693)/(13.11 xx 24 xx 60 xx 60) xx (6.023 xx 10^(23))/(141) xx 0.01`
Total `beta`- particles emitted ` 2.61 xx 10^(23)`
Total energy emittd `= 2.61 xx 10^(13) xx 0.422 = 1.1536 xx 10^(12) MEV`
Energy in erg `= (1.1536 xx 10^(13))(1.6 xx 10^(-6))`
Energy in watt `= (1.536 xx 10^(13) xx 1.6 xx 10^(-6))/(10^(7)) = 1.84` watt


Discussion

No Comment Found

Related InterviewSolutions