JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 22)

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
26

Explanation

Energy released = Change in B.E.

(7.6 $$\times$$ 4) $$-$$ [4 $$\times$$ 1.1] = 26 MeV

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