Assuming the experimental mass of ₆¹²C as 12u, the mass defect of ₆¹²C atom is MeV/c². (Mass of proton = 1.00727u. mass of neutron = 1.00866u, 1u = 931.5MeV/c² and c is the speed of the light in vacuum). (a) 127.5 (b) 89.03 (c) 272.0 (d) 92.0

Answer: (b). ₆¹²C: 6 protons and 6 neutrons. 6(1.00727) + 6(1.00866) = 6.04362 + 6.05196 = 12.09558 u Δ m = 12.09558 - 12 = 0.09558 u = 0.09558 × 931.5 = 89.03 MeV/ c² So the answer is (b) 89.03.

Assuming the experimental mass of ₆¹²C as 12u, the mass defect of… | JEE Main 2026 (5 Apr, Shift 2)

Q3JEE Main 2026 · 5 APR · SHIFT 2Medium

Mass Defect & Binding Energy

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

Show answer and solution

Answer: (b)

${}_{6}^{12}C$ : 6 protons and 6 neutrons.
$6(1.00727) + 6(1.00866) = 6.04362 + 6.05196 = 12.09558$ u
$\Delta m = 12.09558 - 12 = 0.09558$ u $= 0.09558 \times 931.5 = 89.03$ MeV/ $c^2$
So the answer is (b) 89.03.

Report a mistake

Mass Defect & Binding Energy question from JEE Main 2026 (5 Apr, Shift 2)

More questions on Mass Defect & Binding Energy