Question: Q. 5. Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.
Calculate the energy release in $\mathrm{MeV}$ in the deuterium-tritium fusion reaction :
$$ { }{1}^{2} \mathrm{H}+{ }{1}^{3} \mathrm{H} \rightarrow{ }{2}^{4} \mathrm{He}+{ }{0} n^{1} $$
Using the data :
$$ \begin{aligned} & m\left({ }{1}^{2} \mathrm{H}\right)=2.014102 \mathrm{u} \ & m\left({ }{1}^{3} \mathrm{H}\right)=3.016049 \mathrm{u} \end{aligned} $$
$$ \begin{aligned} m\left({ }{2}^{4} \mathrm{He}\right) & =4.002603 \mathrm{u} \ m{n} & =1.008665 \mathrm{u} \ 1 \mathrm{amu} & =931.5 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} $$
U [Delhi I, II, III 2015]
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Solution:
Ans. Nuclear Fission is the breaking down of heavier nucleus into smaller fragments while nuclear fusion is combining of lighter nuclei to form heavier nucleus. We see that binding energy per nucleon of daughter nuclei in both fission and fusion processes is more than that of parent nuclei. Further, the difference in binding energy is released in form of energy while in both the processes certain masses get converted into energy.
$1+1$ In both processes, some mass get converted into energy.
Energy Released
$Q=\left[m\left({ }{1}^{2} \mathrm{H}\right)+m\left({ }{3}^{3} \mathrm{H}\right)-m\left({ }_{2}^{4} \mathrm{He}\right)-m(n)\right]$
$\times 931.5 \mathrm{MeV}$
$=[2.014102+3.016049-4.002603-1.008665]$
- $0,018883 \times 931.5 \mathrm{MeV}$
$\times 931.5 \mathrm{MeV}$
$=17.59 \mathrm{MeV}$
[CBSE Marking Scheme 2015]
