Question

The element curium ${}_{96}^{218}Cm$ has a mean-life of ${10}^{13}s$. Its primary decay modes are spontaneous fission and $\alpha -decay$, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV energy. The masses involved in $\alpha -decay$ are as follows:
${}_{98}^{218}Cm=248.07220u{,}_{94}^{244}Pu=244.064100u$, and ${}_2^{4}He=4.002603u$.
Calculate the power output from a sample of ${10}^{20}Cm$ atoms. $(1u=931MeV{c}^{-2})$

Answer 1

Activity (or rate) of fission:
$A=\frac{dN}{dt}=\lambda N=\frac{1}{\tau }N=\frac{{10}^{20}}{{10}^{13}}={10}^7$

As probability of fission is 8% only, therefore, actual rate of fission is:
$\frac{8}{100}\times {10}^7=8\times {10}^5{s}^{-1}$

Energy released per fission $=200MeV$

Power output of fission $=8\times {10}^5\times 200MeV{s}^{-1}$$=16\times {10}^{7}MeV{s}^{-1}$

Probability of $\alpha -particle$ decay is 92%.

Therefore, rate of decay of $\alpha -particle$ is: $\frac{92}{100}\times {10}^7=92\times {10}^5{s}^{-1}$
For $\alpha -decay$, the equation is:
$\underset{\left(248.072220\right)}{{}_{96}^{248}Cm}\to \underset{\left(244.064100\right)}{{}_{94}^{244}Pu}+\underset{\left(4.002603\right)}{{}_2^{4}He}$

Mass of decay products is:
$244.064100+4.002603=248.06673$.

Mass defect, $\Delta m=24.072220-248.066703=0.005517u$

Energy released per $\alpha -decay$ is:
$92\times {10}^5\times 5/1363MeV{s}^{-1}=4.725\times {10}^{7}MeV{s}^{-1}$

Total power output $=16\times {10}^7+4.725\times {10}^7$
$=20.725\times {10}^{7}MeV{s}^{-1}$
$=20.725\times {10}^7\times 1.6\times {10}^{-13}J{s}^{-1}$
$=33.16\times {10}^{-6}W=33.16\mu W$