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Exam level6 past questions

Mass defect, binding energy, and mass-energy equivalence

Learn how mass defect leads to nuclear binding energy, the use of E=mc², and how to calculate binding energy per nucleon and related quantities.

Why this shows up in the exam

NEET tests your ability to relate mass changes to energy and solve binding energy problems.

How NEET tests this

Numerical · 3 QsDirect recall · 3 Qs

Learn the idea

Mass defect is the difference between the sum of individual nucleon masses and the actual nuclear mass; this missing mass appears as binding energy via E=Δm c², and 1 u corresponds to 931 MeV of energy (i.e., 1 u = 931 MeV/c²).

🧠 Memory hook: Missing mass = hidden power: the ‘lost’ weight of a nucleus fuels the glue that holds it together, just like a secret battery inside a puzzle piece.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Δm = Z Mₚ + N Mₙ – M(N,Z)
  • Binding energy B = Δm c²
  • 1 u = 931.5 MeV/c² (≈931 MeV of energy)
  • Binding energy per nucleon = B/A
  • When atomic masses are used, subtract Z mₑ to obtain nuclear mass
  • c = 2.998×10⁸ m s⁻¹

How to approach it

  1. 1Write the total mass of Z protons and N neutrons using given atomic‑mass units
  2. 2Subtract the actual nuclear mass (or atomic mass minus Z mₑ) to obtain Δm
  3. 3Convert Δm to energy: either multiply by c² or use the factor 931 MeV per u
  4. 4If required, divide by the mass number A to get binding energy per nucleon

Worked example — watch it click

The energy equivalent of one atomic mass unit is:

  • A)1.6 × 10⁻¹⁹J
  • B)6.02 × 10²³ J
  • ✅931 MeV
  • D)9.31 MeV

The concept behind this problem

The example asks for the energy equivalent of 1 u, which is the conversion factor needed whenever we turn a mass defect (in u) into binding energy (in MeV).

Step by step

  1. 1By definition, 1 atomic mass unit (1 u) = 1.66054 × 10⁻²⁷ kg.
  2. 2Using Einstein's mass-energy relation E = mc²: E = (1.66054 × 10⁻²⁷ kg)(2.998 × 10⁸ m/s)² = 1.4924 × 10⁻¹⁰ J.
  3. 3Converting to eV: E = 1.4924 × 10⁻¹⁰ J / (1.602 × 10⁻¹⁹ J/eV) ≈ 931.5 MeV ≈ 931 MeV.
  4. 4This is a standard conversion factor used in nuclear physics.

Watch out

Students often write ‘1 u = 931 MeV’ without the essential “/c²”, treating a mass unit as an energy unit.

Common slip-ups that cost marks

  • •For atomic masses, forget to remove the electron masses of Z atoms
  • •Using 931 MeV as a mass unit instead of 931 MeV/c²
  • •Rounding the conversion factor too early, which spoils the final answer

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 6NEET 2003

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of ₂He⁴ is:

Push further

More challenging

3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 3

A deuterium nucleus (₁H²) has a mass of 2.014102 u. Given that the mass of a proton is 1.007825 u and the mass of a neutron is 1.008665 u, calculate the binding energy of the deuterium nucleus. (Take 1 u = 931.5 MeV/c²)

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