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

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Modern Physics

The binding energy per nucleon of

_{83}^{209} Bi

____

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✓ Correct
(C)8.79
(D)6.94

Answer: (B)

Solution

The mass defect of the nucleus is

Δ m = 83 m_{p} + 126 m_{n} - M_{B i} = 83 (1.007825) + 126 (1.008665) - 208.980388 = 1.760877 u

. The total binding energy is

Δ m \times 931 = 1.760877 \times 931 = 1639.4 MeV

. Dividing by the

209

nucleons, the binding energy per nucleon is

\frac{1639.4}{209} = 7.84 MeV

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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