JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Modern Physics

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Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are

5.6

MeV,

7.4

MeV and

6.1

MeV, respectively, then in the process,

Δ M c^{2} =

______ MeV.

(A) $6.9$
(B) $7.9$
(C) $2.2$ ✓ Correct
(D) $4.3$

Answer: (C)

Solution

The energy change of a nuclear reaction equals the total binding energy of the products minus that of the reactants. Here

BE_{products} = 10 \times 6.1 = 61

MeV and

BE_{reactants} = 2 (3 \times 5.6) + (4 \times 7.4) = 33.6 + 29.6 = 63.2

MeV. Thus

Δ M c^{2} = 61 - 63.2 = - 2.2

MeV, so the magnitude of the energy involved in the process is

2.2

MeV.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option

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