JEE Main 2023 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 13 April 2023, Shift 1, Physics Q46

Question 46 of 90 in this shift, Physics question 16 of 30, Section A.

^{238}_{92}A \rightarrow\, ^{234}_{90}B + \,^{4}_{2}D + Q

In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of

^{238}_{92}A = 238{\cdot}05079 \times 931.5\ \mathrm{MeV/c^2}

, mass of

^{234}_{90}B = 234{\cdot}04363 \times 931{\cdot}5\ \mathrm{MeV/c^2}

, mass of

^{4}_{2}D = 4{\cdot}00260 \times 931{\cdot}5\ \mathrm{MeV/c^2}

]

(1) $3{\cdot}82$ MeV
(2) $4{\cdot}25$ MeVOfficial answer
(3) $5{\cdot}9$ MeV
(4) $2{\cdot}12$ MeV

Official answer

Option 2

NTA final key.

Chapter: Nuclei
Topic: Nuclear Binding Energy
Idea tested: Alpha Decay Q-Value with Atomic Masses

Same topic in other shifts

All Nuclear Binding Energy questions

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 108 of the 174 JEE Main shifts held from 2017 to 2026.