Assuming the experimental mass of 12 6 C as 12 u the mass defect of 12 6 C atom is MeV/c². (Mass of proton = 1.00727u, mass of neutron = 1.00866u, 1 u = 931.5 MeV /c 2 )

Physics · Nuclear Physics

JEE Main 2026 — 5 April, Evening Shift — Question 18

Assuming the experimental mass of $^{12}_{6}\mathrm{C}$ as $12\mathrm{u}$ the mass defect of $^{12}_{6}\mathrm{C}$ atom is MeV/c². (Mass of proton = 1.00727u, mass of neutron = 1.00866u, $1\mathrm{u} = 931.5\mathrm{MeV}/c^2$ )

Option A:
127.5
Option B:
89.03
Correct
Option C:
272
Option D:
92

Answer: B

Step-by-step solution

$\Delta m = (6\times1.00727 + 6\times1.00866) - 12 = 0.09558$ u. Energy = $0.09558 \times 931.5 = 89.03$ MeV/c².

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction