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

Physics · Nuclear Physics

JEE Main 2026 — 2 April, Evening Shift — Question 16

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

Option A:
7.48
Option B:
7.84
Correct
Option C:
8.79
Option D:
6.94

Answer: B

Step-by-step solution

$\Delta m = 83m_p+126m_n - m_{Bi}=1.760877u$ , BE = 1.760877×931=1639.3764 MeV, BE/A = 1639.3764/209=7.8439 MeV

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction