CBSE 2025 · Region 6 · Set 1 · Q28 · 3 marks
(a)Consider the so-called'D-T reaction' (Deuterium-Tritium reaction) . In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}+\mathrm{Q} \] Find the amount of energy released in the reaction. Given: \[\begin{aligned} & \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)=2 \cdot 014102 \mathrm{u} \\ & \mathrm{~m}\left({ }_{1}^{3} \mathrm{H}\right)=3 \cdot 016049 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4 \cdot 002603 \mathrm{u} \\ & \mathrm{~m}\left({ }_{0}^{1} \mathrm{n}\right)=1 \cdot 008665 \mathrm{u} \\ & 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} \](b)Show that the nuclear density is independent of mass number.
(a)
Consider the so-called'D-T reaction' (Deuterium-Tritium reaction) . In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}+\mathrm{Q} \] Find the amount of energy released in the reaction. Given: \[\begin{aligned} & \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)=2 \cdot 014102 \mathrm{u} \\ & \mathrm{~m}\left({ }_{1}^{3} \mathrm{H}\right)=3 \cdot 016049 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4 \cdot 002603 \mathrm{u} \\ & \mathrm{~m}\left({ }_{0}^{1} \mathrm{n}\right)=1 \cdot 008665 \mathrm{u} \\ & 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} \]
(b)
Show that the nuclear density is independent of mass number.
Marking-scheme solution
| Finding the amount of energy released | $\displaystyle 2$ | |
| Showing the nuclear density is independent of mass number | $\displaystyle 1$ | |
| \[ \begin{aligned} \Delta \mathrm{m} | =\left[\mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)+\mathrm{m}\left({ }_{1}^{3} \mathrm{H}\right)\right]-\left[\mathrm{m}\left({ }_{2}^{4} \mathrm{H} e\right)+\mathrm{m}\left({ }_{0}^{1} \mathrm{n}\right)\right] | |
| =(2.014102+3.016049)-(4.002603+1.008665) | ||
| =0.018883 \mathrm{u} \end{aligned} \] \[ \begin{aligned} \mathrm{Q}= | \Delta \mathrm{m} \times 931 | |
| = | 0.018883 \times 931 \mathrm{MeV} | |
| \mathrm{Q}=17.58 \mathrm{MeV} \end{aligned} \] | ||
| Nuclear density \(\displaystyle =\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}\) \[ \rho=\frac{\mathrm{m} A}{\dfrac{4}{3} \pi R^{3}} \] \[ \begin{aligned} | R=R_{0} A^{1 / 3} | |
| \rho=\frac{3 \mathrm{m}}{4 \pi R_{0}^{3}} \end{aligned} \] | ||
| Independent of mass number (A) |
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.