CBSE 2024 · Region 5 · Set 2 · Q20 · 2 marks
Calculate the energy released/absorbed (in MeV ) in the nuclear reaction : \[\begin{aligned} &{ }_{1}^{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \\ & \text { Given : } \mathrm{m}\left({ }_{1}^{1} \mathrm{H}\right)=1.007825 \mu \\ & \mathrm{~m}\left({ }_{1}^{2} \mathrm{H}\right)=2.014102 \mu \\ & \mathrm{~m}\left({ }_{1}^{3} \mathrm{H}\right)=3.016049 \mu \end{aligned} \]
Marking-scheme solution
Mass of reactants $\displaystyle =(1.007825+3.016049) \mathrm{u}$
$\displaystyle =4.023874 \mathrm{u}$
Mass of product $\displaystyle =2 \times 2.014102 \mathrm{u}$
$\displaystyle =4.028204 \mathrm{u}$
Mass defect, $\displaystyle \Delta m=4.023874 \mathrm{u}-4.028204 \mathrm{u}$
$\displaystyle =-0.00433 \mathrm{u}$
As the mass defect is negative, energy is absorbed.
Energy absorbed, $\displaystyle E=0.00433 \times 931.5 \mathrm{MeV}$
$\displaystyle =4.03 \mathrm{MeV}$
NucleiMass-Energy and Nuclear Binding EnergyApplyvery_short_answermedium
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