CBSE 2024 · Region 5 · Set 3 · Q20 · 2 marks
Deuterium undergoes the following fusion reaction : \[{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}+3.27 \mathrm{MeV} \] How long an electric bulb of $\displaystyle 200$ W will glow by using the energy released in $\displaystyle 2$ g of deuterium?
Marking-scheme solution
Number of atoms in $\displaystyle 2$ g deuterium $\displaystyle =6.023 \times 10^{23}$
Energy released per atom $\displaystyle =\frac{3.27}{2}=1.635 \mathrm{MeV}$
$\displaystyle t=\frac{\text { Total energy released }}{\text { Power }}$
$\displaystyle t=\frac{6.023 \times 10^{23} \times 1.635 \times 1.6 \times 10^{-13}}{200}$
$\displaystyle t=7.88 \times 10^{8} \mathrm{~s}$
NucleiMass-Energy and Nuclear Binding EnergyApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.