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

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Modern Physics

Assuming the experimental mass of

_{6}^{12} C

as 12 u, the mass defect of

_{6}^{12} C

atom is ______

MeV / c^{2}

. (Mass of proton

= 1.00727 u

. mass of neutron

= 1.00866 u, 1 u = 931.5 MeV / c^{2}

and c is the speed of the light in vacuum.)

(A)127.5
(B)89.03✓ Correct
(C)272.0
(D)92.0

Answer: (B)

Solution

The mass defect is the amount by which the summed mass of the free nucleons exceeds the measured mass of the atom,

Δ m = (Z m_{p} + N m_{n}) - m

. For

_{6}^{12} C

with

Z = 6

and

N = 6

Δ m = 6 (1.00727) + 6 (1.00866) - 12 = 6.04362 + 6.05196 - 12 = 0.09558 u

. Converting with

1 u = 931.5 MeV / c^{2}

Δ m = 0.09558 \times 931.5 = 89.03 MeV / c^{2}

. The mass defect of the carbon-12 atom is

89.03 MeV / c^{2}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

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