JEE Main 2026 · 6 Apr · Shift 1 · Physics Q48

← JEE Main 2026 · 6 Apr · Shift 1

PhysicsInteger+4 / -1Question 48 of 75

The energy released when $\dfrac7{17.13}$ 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/c² and Avogadro number = $6.0\times10^{23}$ )

Your answer

Answer and worked solution

Official answer: 6

Reaction. $\ce{^7_3Li + ^1_1H -> 2^4_2He}$ .

Energy per reaction.
$\Delta m=7.0183+1.008-8.008=0.0183\text{ u},$
which is $0.0183\times931\approx17.04$ MeV.

Number of nuclei.
$\frac{7000/17.13}7\times6\times10^{23}=3.50\times10^{25}.$

Total. $3.50\times10^{25}\times17.04\times10^6\text{ eV}\approx6\times10^{32}$ eV.

Answer: 6

Topics:Mass defect and binding energy Nuclear fission and fusion

← Previous question Next question →

JEE Main 2026 · 6 Apr · Shift 1, Physics question 48