Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :
${ }_{1}^{2} X+{ }_{1}^{2} X={ }_{2}^{4} Y$
The binding energies per nucleon for $\frac{2}{1} X$ and ${ }_{2}^{4} Y$ are $1.1 \,\mathrm{MeV}$ and $7.6 \,\mathrm{MeV}$ respectively. The energy released in this process is _______________ $\mathrm{MeV}$.
Answer (integer)
26
Solution
<p>Energy released = Change in B.E.</p>
<p>(7.6 $\times$ 4) $-$ [4 $\times$ 1.1] = 26 MeV</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Fission and Fusion
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