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Physics · 2022 · 3 marks
CBSE 2022 · Region 4 · Set 1 · Q4
(a) Differentiate between nuclear fission and nuclear fusion. (b) Deuterium undergoes fusion as per the reaction: _1^2 H +_1^2 H →_2^3 He +_0^1 n +3…
(a)
Differentiate between nuclear fission and nuclear fusion.
(b)
Deuterium undergoes fusion as per the reaction: \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}+3 \cdot 27 \mathrm{MeV} \] Find the duration for which an electric bulb of $\displaystyle 500$ W can be kept glowing by the fusion of $\displaystyle 100$ g of deuterium.
Marking-scheme solution
a) Nuclear fission - The process of breaking a very heavy nucleus into lighter nuclei, having mass number in the range of middle mass number ($\displaystyle 30$ < A < $\displaystyle 170$). Nuclear Fusion- It is the process of joining of very light nuclei $\displaystyle (\mathrm{A} \leq 10)$ to form a heavier nucleus.
b) No. of atoms in $\displaystyle 100 \mathrm{~g}=\frac{6.023 \times 10^{23}}{2} \times 100=3.0115 \times 10^{25}$ Energy released/atom $\displaystyle =\frac{3.27 \mathrm{MeV}}{2}=1.635 \mathrm{MeV}$
Total energy released $\displaystyle =3.0115 \times 10^{25} \times 1.635 \mathrm{MeV}$
$$\begin{aligned}
& =3.0115 \times 1.635 \times 10^{25} \times 1.6 \times 10^{-13} \\
& =7.878 \times 10^{12} \mathrm{~J}
\end{aligned}
$$
$$\begin{aligned}
t & =\frac{E}{P} \\
& =\frac{7 \cdot 878 \times 10^{12} \mathrm{~J}}{500 \mathrm{~J} / \mathrm{s}}=1.5756 \times 10^{10} \mathrm{~s} \\
& =\frac{1.5756 \times 10^{10}}{3 \cdot 15 \times 10^{7}} \simeq 500 y
\end{aligned}
$$
Alternatively :
$$\begin{aligned}
& E=\frac{M Q}{2 m_{d}} \\
& =\frac{(0 \cdot 1 \mathrm{~kg}) \times(3 \cdot 27 \mathrm{MeV})}{2(2 \cdot 04) \times\left(1 \cdot 66 \times 10^{-27} \mathrm{~kg} / 4\right)} \\
& =0 \cdot 0492 \times 10^{27}=4 \cdot 92 \times 10^{25} \mathrm{MeV} \\
& t=\frac{E}{P} \\
& =\frac{\left(4 \cdot 92 \times 10^{25}\right) \times\left(1 \cdot 6 \times 10^{-13}\right)}{500}=1.5756 \times 10^{10} \mathrm{~s} \\
& =\frac{1.5756 \times 10^{10}}{3 \cdot 15 \times 10^{7}} \simeq 500 y
\end{aligned}
$$More from Nuclei
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CBSE Class 12 Physics past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.