Question

The element curium ${}_{96}{\text{Cm}}^{248}$ has a mean life of ${10}^{13}\text{s}$. Its primary decay modes are spontaneously fission and $\alpha$ decay, the former with a probability of $8\%$ and the latter with a probability of $92\%$. Each fission releases $200\text{MeV}$ of energy. The masses involved $\alpha$ decay are as follows:
${}_{96}{\text{Cm}}^{248}=248.072220u,{}_{94}{\text{Pu}}^{244}=244.064100u{}^{}{}_2{\text{He}}^4=4.002603u.$
Calculate the power output from a sample of ${10}^{20}$ $\text{Cm}$ atoms $(1u=931\frac{\text{MeV}}{c^2})$.

• A. $3.3\times {10}^{-5}\text{W}$
• B. $3.3\times {10}^5\text{W}$
• C. $8.8\times {10}^{-5}\text{W}$
• D. $1.1\times {10}^5\text{W}$

Answer 1

The correct option is A $3.3\times {10}^{-5}\text{W}$

The $\alpha$-decay is represented as,

${}_{96}{\text{Cm}}^{248}\to {}_{94}{\text{Pu}}^{244}+{}_2{\text{He}}^4$

The mass defect
$\Delta m=248.072220u-(244.064100u+4.002603u)$

$\Rightarrow \Delta m=0.005517u$

$\text{Equivalent energy},$

$E_{\alpha }=0.005517\times 931=5.136\text{MeV}$

Energy released in the decay of one atom is

$E=E_{\alpha }+E_{fission}$

Given that, $\alpha$ decay has $92\%$ probability and fission has $8\%$ probability

$\Rightarrow E=0.92\times 5.136+0.08\times 200$

$\Rightarrow E=20.725\text{MeV}$

$\text{Total energy released in}{10}^{20}\text{atoms}$

$E_T={10}^{20}\times 20.725\text{MeV}$

$\text{Power output}=\frac{\text{Total energy released}}{\text{Mean life}}$

$\Rightarrow P=\frac{{10}^{20}\times 20.725\times 1.6\times {10}^{-13}}{{10}^{13}}$

$\Rightarrow P=3.3\times {10}^{-5}\text{W}$

Hence, option $\left(A\right)$ is correct.