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CBSE 2022 · Region 1 · Set 1 · Q6 · 3 marks

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 : $\displaystyle \mathrm{m}\left({ }_{1}^{2} \mathrm{H}\right)=2.014102 \mathrm{u}$ \[\begin{aligned} & \mathrm{m}\left({ }_{1}^{3} \mathrm{H}\right)=3.016049 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u} \\ & \mathrm{~m}_{\mathrm{n}}=1.008665 \mathrm{u} \end{aligned} \]

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