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Physics · 2022 · 3 marks
CBSE 2022 · Region 1 · Set 1 · Q6
Calculate the energy released in MeV in the following reaction: _1^2 H +_1^3 H →_2^4 He + n Given: m (_1^2 H )=2.014102 u m (_1^3 H )=3.016049 u; m…
Calculate the energy released in MeV in the following reaction : \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n} \] Given : $\displaystyle \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)=2.014102 \mathrm{u}$ \[\begin{aligned} & \mathrm{m}\left({ }_{1}^{3} \mathrm{H}\right)=3.016049 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u} \\ & \mathrm{~m}_{\mathrm{n}}=1.008665 \mathrm{u} \end{aligned} \]
Marking-scheme solution
$${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+n+\text { Energy }
$$
Mass defect = mass of reactants - mass of products
$$\Delta m=m\left({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H}\right)-m\left({ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} n\right)
$$
Mass defect $\displaystyle =(2 \cdot 014102+3 \cdot 016049)-(4 \cdot 002603+1 \cdot 008665)$
=$\displaystyle 5.030151$-$\displaystyle 5.011268$
= 0.018883u
Energy released $\displaystyle =\Delta \mathrm{m} \times 931.5 \mathrm{MeV}$
$$=0.018883 \times 931.5 \mathrm{MeV}
$$
$$=17.58 \mathrm{MeV}
$$
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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.