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Friday, 9 October

JEE Main 2025 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2025, Shift 2, Physics Q38

Question 38 of 75 in this shift, Physics question 13 of 25, Section A.

Energy released when two deuterons (1H2_1\mathrm{H}^2) fuse to form a helium nucleus (2He4_2\mathrm{He}^4) is : (Given : Binding energy per nucleon of 1H2=1.1_1\mathrm{H}^2 = 1.1 MeV and binding energy per nucleon of 2He4=7.0_2\mathrm{He}^4 = 7.0 MeV)
  1. (1)8.18.1 MeV
  2. (2)5.95.9 MeV
  3. (3)23.623.6 MeVOfficial answer
  4. (4)26.826.8 MeV

Official answer

Option 3

NTA final key.

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

Same idea in other shifts

Asked 2× in all
  1. 2 Apr 2026, Shift 2 · Q44The binding energy per nucleon of 20983Bi\frac{209}{83}Bi isMediumSingle 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.