Assuming the experimental mass of 12 6C as 12 u, the mass defect of 12 6C 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).

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q44 – JEE-MAIN 2026

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Home Physics NucleiQuestion #367

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).

Clue 1 of 3: Nuclear Structure

Understand that the mass defect arises from the difference between the mass of the constituent nucleons (protons and neutrons) and the actual mass of the nucleus.

🧠 Full Solution Path

Step 1: Identify Constituents and Calculate Theoretical Mass✦ Active

For a $_{6}^{12} C$ atom, there are 6 protons (Z=6) and 6 neutrons (N=12-6=6). The theoretical mass of the atom is the sum of the masses of its constituent particles. Assuming the given 'mass of proton' includes the electron, it represents the mass of a hydrogen atom ( $m_{H}$ ). Therefore, the theoretical mass ( $M_{t h e o}$ ) is:

M_{t h e o} = (6 \times m_{H}) + (6 \times m_{n})

M_{t h e o} = (6 \times 1.00727 u) + (6 \times 1.00866 u)

M_{t h e o} = 6 \times (1.00727 + 1.00866) u

M_{t h e o} = 6 \times (2.01593) u

M_{t h e o} = 12.09558 u

Step 2: Calculate Mass Defect○ Expand

The mass defect ( $Δ m$ ) is the difference between the theoretical mass and the experimental mass of the atom. The experimental mass ( $M_{e x p}$ ) is given as 12 u.

Δ m = M_{t h e o} - M_{e x p}

Δ m = 12.09558 u - 12 u

Δ m = 0.09558 u

Step 3: Convert Mass Defect to MeV/c

^{2}

○ Expand

Convert the mass defect from atomic mass units (u) to MeV/c $^{2}$ using the given conversion factor: $1 u = 931.5 {MeV/c}^{2}$ .

Δ m = 0.09558 u \times 931.5 \frac{MeV}{c^{2} \cdot u}

Δ m = 89.03007 \frac{MeV}{c^{2}}

Δ m \approx 89.03 \frac{MeV}{c^{2}}

This value matches option 2.

💡 Teacher's Secret Hint

Ensure correct unit conversion and significant figures for the final answer.

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: Nuclear Structure

🧠 Full Solution Path