JEE Main 2026 · 2 Apr · Shift 2 · Physics Q44

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PhysicsSingle correct+4 / -1Question 44 of 75

The binding energy per nucleon of ${}^{209}_{83}Bi$ is ________ MeV.
[Take $m\left({}^{209}_{83}Bi\right)=208.980388$ u, $m_p=1.007825$ u, $m_n=1.008665$ u, 1 u = 931 MeV/c²]

7.48

7.84

8.79

6.94

Answer and worked solution

Official answer: B

Mass defect.
$\Delta m=83(1.007825)+126(1.008665)-208.980388=1.760877\text{ u}.$

Binding energy. $1.760877\times931=1639.4$ MeV; per nucleon $\frac{1639.4}{209}=7.84$ MeV.

(B)

Topics:Mass defect and binding energy

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JEE Main 2026 · 2 Apr · Shift 2, Physics question 44