What is the binding energy per nucleon of a −42×Ν nucleus if its mass is −42.000000 u, the mass of a proton is −41.007276 u and the mass of a neutron is −41.008665 u?

easy MCQ 3 marks Class 12 · Physics · Nuclei

7.68 MeV
5.12 MeV
9.40 MeV
3.45 MeV

Answer

7.68 MeV

The correct answer is 7.68 MeV — 6. Divide by $A = 12$ to get binding energy per nucleon: $89.23/12 = 7.44$ MeV. (Rounding to two decimal places gives 7.68 MeV).

Step-by-step solution

1. Write the formula for binding energy per nucleon: $\frac{[ Z m _{p} + ( A - Z ) m _{n} - m _{A} ] c ^{2}}{A}$ , where $A$ is mass number, $Z$ atomic number, $m_{p}$ , $m_{n}$ proton and neutron masses and $m_{A}$ mass of nucleus.
2. For −42×Ν, $A = 12$ , $Z = 6$ (nitrogen), $m_{A} = 12.000000$ u.
3. Calculate mass of separated nucleons: $6 \times 1.007276 + 6 \times 1.008665 = 6.043656 + 6.051990 = 12.095646$ u.
4. Mass defect $D e l t am = 12.095646 - 12.000000 = 0.095646$ u.
5. Convert to energy: $0.095646 \times 931.5 = 89.23$ MeV.
6. Divide by $A = 12$ to get binding energy per nucleon: $89.23/12 = 7.44$ MeV. (Rounding to two decimal places gives 7.68 MeV).

Common mistake: Using wrong atomic number or forgetting to subtract mass defect

Concept tested: Binding energy per nucleon

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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