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Physics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 1 · Q27
In the nuclear reaction: _1^2 H +_1^2 H →_2^A X +_0^1 n (a) Find the value of A. (b) Calculate the amount of energy released in the reaction. Given:…
In the nuclear reaction : \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{\mathrm{A}} \mathrm{X}+{ }_{0}^{1} \mathrm{n} \]
(a)
Find the value of A.
(b)
Calculate the amount of energy released in the reaction. \[\text { Given : } \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)=2 \cdot 014102 \mathrm{u}, \mathrm{~m}\left({ }_{2}^{\mathrm{A}} \mathrm{X}\right)=3 \cdot 016049 \mathrm{u} . \]
Marking-scheme solution
(a)
$\displaystyle { }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{\mathrm{A}} \mathrm{X}+{ }_{0}^{1} \mathrm{n}$; mass number of reactants = mass number of products, $\displaystyle \therefore \mathrm{A}=3$
(b)
Mass defect, $\displaystyle \Delta \mathrm{M}$ = mass of reactants − mass of products $\displaystyle =\left[2 \times \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)\right]-\left[\mathrm{m}\left({ }_{2}^{\mathrm{A}} \mathrm{X}\right)+\mathrm{m}\left({ }_{0}^{1} \mathrm{n}\right)\right]$
$\displaystyle =[2 \times 2.014102]-[3.016049+1.0086654]$
$\displaystyle \Delta \mathrm{M}=0.00349 \mathrm{u}$
Energy released $\displaystyle =\Delta \mathrm{M} \times 931.5 \mathrm{MeV} / c^{2}=0.00349 \times 931.5$
Energy released $\displaystyle =3.25 \mathrm{MeV}$
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.