The binding energy per nucleon of $\frac{209}{83}Bi$ is

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumRecall

JEE Main 2 April 2026, Shift 2, Physics Q44

Question 44 of 75 in this shift, Physics question 19 of 25, Section A.

The binding energy per nucleon of

\frac{209}{83}Bi

(1) $7.48$ MeV
(2) $7.84$ MeVOfficial answer
(3) $8.79$ MeV
(4) $6.94$ MeV

Official answer

Option 2

NTA final key.

Chapter: Nuclei
Topic: Nuclear Binding Energy
Idea tested: Binding Energy per Nucleon

Same idea in other shifts

Asked 2× in all

2 Apr 2025, Shift 2 · Q38Energy released when two deuterons ( $_1\mathrm{H}^2$ ) fuse to form a helium nucleus ( $_2\mathrm{He}^4$ ) is : (Given : Binding energy per…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.