/

Learn

Saturday, 31 October

/

JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 4 April 2024, Shift 2, Physics Q51

Question 51 of 90 in this shift, Physics question 21 of 30, Section B.

The disintegration energy

Q

for the nuclear fission of

^{235}U \to {}^{140}Ce + {}^{94}Zr + n

is ________ MeV. Given atomic masses of

^{235}U

: 235.0439

u

;

^{140}Ce

: 139.9054

u

,

^{94}Zr

: 93.9063

u

;

n

: 1.0086

u

, Value of

c^2

= 931

MeV/u

.

Official answer

208

NTA final key.

Chapter: Nuclei
Topic: Fission and Fusion
Idea tested: Q-Value of a Reaction

Same idea in other shifts

Asked 2× in all

9 Apr 2024, Shift 1 · Q60A star has $100\%$ helium composition. It starts to convert three ${}^4\mathrm{He}$ into one ${}^{12}\mathrm{C}$ via triple alpha process…HardNumerical value

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.