JEE Main 2026 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 6 April 2026, Shift 1, Physics Q48

Question 48 of 75 in this shift, Physics question 23 of 25, Section B.

The energy released when

\frac{7}{17.13}

kg of

^{7}_{3}\mathrm{Li}

is converted into

^{4}_{2}\mathrm{He}

by proton bombardment is

\alpha\times10^{32}

eV. The value of

\alpha

is ________. (Nearest integer) (Mass of

^{7}_{3}\mathrm{Li}=7.0183

u, mass of

^{4}_{2}\mathrm{He}=4.004

u, mass of proton

=1.008

u and 1 u

=931

MeV/c

^2

and Avogadro number

=6.0\times10^{23}

)

Official answer

NTA final key.

Chapter: Nuclei
Topic: Nuclear Binding Energy
Idea tested: Macroscopic Energy Yield

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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.