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
Comments (0)
