JEE Main 2026 Apr 4 Shift 1, Physics Q4: Mass-Energy and Nuclear Binding Energy

JEE Main2026Apr 4, Shift 1Physics

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
$2.2$
D
$4.3$

Solution

The energy released (mass defect $\times c^{2}$ ) equals the increase in total binding energy from reactants to product.

BE of reactants: $2 (3 \times 5.6) + (4 \times 7.4) = 33.6 + 29.6 = 63.2$ MeV.

BE of product: $10 \times 6.1 = 61.0$ MeV.

$Δ M c^{2} = BE_{reactants} - BE_{product} = 63.2 - 61.0 = 2.2$ MeV.

Concept behind this question

Mass-Energy and Nuclear Binding EnergyNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 4 Shift 1, Physics Q4: Mass-Energy and Nuclear Binding Energy

Concept behind this question

More previous year questions on Mass-Energy and Nuclear Binding Energy

Practice more Physics