JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

The binding energy per nucleon of $_{83}^{209} Bi$ is _______ MeV. [Take $m (_{83}^{209} Bi) = 208.980388$ u, $m_{p} = 1.007825$ u, $m_{n} = 1.008665$ u, $1$ u $= 931$ MeV/ $c^{2}$ ]

A
$7.48$
B
$7.84$
C
$8.79$
D
$6.94$

Solution

The nucleus has $83$ protons and $209 - 83 = 126$ neutrons. Mass defect:

$Δ m = 83 \cdot 1.007825 + 126 \cdot 1.008665 - 208.980388 = 83.649475 + 127.091790 - 208.980388 = 1.760877$ u.

Total binding energy: $BE = 1.760877 \cdot 931 \approx 1639.38$ MeV.

Per nucleon: $\frac{BE}{A} = \frac{1639.38}{209} \approx 7.84$ MeV.

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