Q69 / 75 · JEE Main 2026 (05 April Shift 2) · Physics
Assuming the experimental mass of ${}^{12}_{6}C$ as $12\text{ u}$, the mass defect of ${}^{12}_{6}C$ 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).
(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).
- A$127.5$
- Correct answerB$89.03$
- C$272.0$
- D$92.0$
Solution
The mass of $6$ protons is $6 \times 1.00727\text{ u} = 6.04362\text{ u}$.
The mass of $6$ neutrons is $6 \times 1.00866\text{ u} = 6.05196\text{ u}$.
Total mass of the constituent nucleons is $6.04362\text{ u} + 6.05196\text{ u} = 12.09558\text{ u}$.
The mass defect $\Delta m$ is the difference between the total mass of the nucleons and the experimental mass of the atom (ignoring electron mass as per the given data):
$\Delta m = 12.09558\text{ u} - 12\text{ u} = 0.09558\text{ u}$
Converting the mass defect into $\text{MeV}/c^2$:
$\Delta m = 0.09558 \times 931.5\text{ MeV}/c^2 = 89.03277\text{ MeV}/c^2 \approx 89.03\text{ MeV}/c^2$
Answer: $89.03$
The mass of $6$ neutrons is $6 \times 1.00866\text{ u} = 6.05196\text{ u}$.
Total mass of the constituent nucleons is $6.04362\text{ u} + 6.05196\text{ u} = 12.09558\text{ u}$.
The mass defect $\Delta m$ is the difference between the total mass of the nucleons and the experimental mass of the atom (ignoring electron mass as per the given data):
$\Delta m = 12.09558\text{ u} - 12\text{ u} = 0.09558\text{ u}$
Converting the mass defect into $\text{MeV}/c^2$:
$\Delta m = 0.09558 \times 931.5\text{ MeV}/c^2 = 89.03277\text{ MeV}/c^2 \approx 89.03\text{ MeV}/c^2$
Answer: $89.03$
Hence, the correct answer is option: 'B'.