The radioactive disintegration of `""_(94)^(239)Pu and alpha`-emission process is accompanied by loss of 5.24 M ev/dis. If `t_(0.5)` of `""_(94)^(239)Pu` is `2.44 xx 10^(4)` year, calculate the energy released per year from 1.0 g sample of `""_(94)^(239)Pu` in kJ.

Options

A. 52.62 kJ
B. 60.06 kJ
C. 56.73 kJ
D. 68.74 kJ

Answer

B

Explanation

Rate ` = lambdaN (0.693)/(2.44 xx 10^(4)) xx (6.02 xx 10^(23))/(239) = 7.153 xx 10^(16)` dis/year
loss in energy per year in emission of `alpha`-particles` = 5.24 xx 7.153 xx 10^(6)` Mev
` = 37.482 xx 10^(16) xx 1.602 xx 10^(-16)kJ = 60.05 kJ`