Modern Physics 5 Question 32
35. The element curium ${ } _{96}^{248} Cm$ 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 $200 MeV$ of energy. The masses involved in decay are as follows :
$(1997,5$ M)
${ } _{96}^{248} Cm=248.072220 u$,
${ } _{94}^{244} Pu=244.064100 u$ and ${ } _2^{4} He=4.002603 u$. Calculate the power output from a sample of $10^{20} Cm$ atoms.
$$ \left(1 u=931 MeV / c^{2}\right) $$
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Solution:
- The reaction involved in $\alpha$-decay is
$$ { } _{96}^{248} Cm \rightarrow{ } _{94}^{244} Pu+{ } _2^{4} He $$
Mass defect
$\Delta m=$ mass of ${ } _{96}^{248} Cm-\operatorname{mass}$ of ${ } _{94}^{244} Pu-$ mass of ${ } _2^{4} He$
$=(248.072220-244.064100-4.002603) u$
$=0.005517 u$
Therefore, energy released in $\alpha$-decay will be
$$ E _{\alpha}=(0.005517 \times 931) MeV=5.136 MeV $$
Similarly, $E _{\text {fission }}=200 MeV$ (given)
Mean life is given as $t _{\text {mean }}=10^{13} s=1 / \lambda$
$\therefore$ Disintegration constant $\lambda=10^{-13} s^{-1}$
Rate of decay at the moment when number of nuclei are $10^{20}$
$$ \begin{aligned} & =\lambda N=\left(10^{-13}\right)\left(10^{20}\right) \\ & =10^{7} \text { disintegration per second } \end{aligned} $$
Of these disintegrations, $8 %$ are in fission and $92 %$ are in $\alpha$-decay.
Therefore, energy released per second
$$ \begin{aligned} & =\left(0.08 \times 10^{7} \times 200+0.92 \times 10^{7} \times 5.136\right) MeV \\ & =2.074 \times 10^{8} MeV \end{aligned} $$
$\therefore$ Power output (in watt)
$$ \begin{aligned} & =\text { energy released per second }(J / s) \\ & =\left(2.074 \times 10^{8}\right)\left(1.6 \times 10^{-13}\right) \end{aligned} $$
$\therefore$ Power output $=3.32 \times 10^{-5} W$
