Definition
Mass defect and binding energy
Mass defect Δm = Z m_p + N m_n - m_nucleus (always positive). Binding energy B = Δm c². Binding energy per nucleon B/A peaks around A = 56-60 (Fe) at ~8.8 MeV.
-- NCERT Class 12 Physics, Ch. 13, p. 311Binding Energy Variation
8 MCQs1 revision card9-step worked example
Source: NCERT Atoms and NucleiOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Binding Energy Variation, explained for NEET
The curve of binding energy per nucleon against mass number is read wrongly more often than it is read at all. The commonest error: treating total binding energy as the y-axis. It is not. The plot is B/A — binding energy divided by nucleon count. Total B rises almost monotonically with A right up to uranium; B/A does not. Confusing the two makes the whole shape meaningless.
Read the curve in three regions. Below about A = 20 it climbs steeply and erratically, with sharp local peaks at ⁴He, ¹²C and ¹⁶O — nuclei whose nucleon numbers are multiples of four. From roughly A = 30 to A = 170 the curve is nearly flat at 8.0–8.8 MeV per nucleon, peaking near A = 56 (the iron region) at about 8.8 MeV. Beyond A ≈ 170 it falls slowly, reaching about 7.6 MeV per nucleon at ²³⁸U.
Two features carry the physics. The flatness of the middle region means a nucleon is bound to its immediate neighbours only — if every nucleon attracted every other, B/A would grow with A, not plateau. This is the saturation of the nuclear force (NCERT Class 12 Physics, Chapter 13, page 313). The fall-off at high A is Coulomb repulsion between protons, which is long-range and therefore does grow with proton count.
The bridge to NEET: because the middle is highest, moving toward A ≈ 56 from either end increases B/A and releases energy. That single sentence is the whole energetic basis of fission and fusion, and it is what NEET asks you to infer from the shape.
Watch out for the sign. Binding energy is quoted positive, but a more tightly bound nucleus sits lower in total energy. A high point on the B/A curve is a stable nucleus, not an energetic one.
Can you answer these Binding Energy Variation MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the standard plot used to discuss nuclear stability, the quantity on the vertical axis is
Show answer and why every option is right or wrong
Answer: B. The curve plots B/A against A. NCERT Class 12 Physics, Chapter 13, page 312, introduces it as the binding energy per nucleon curve.
Why A is wrong: A is wrong because total binding energy rises almost steadily with A all the way to uranium and shows none of the plateau-and-fall shape the curve is famous for.
Why C is wrong: C is wrong because mass defect, like total B, grows with nucleus size; dividing by A is exactly what produces the characteristic shape.
Why D is wrong: D is wrong because A is the horizontal axis, not the vertical one.
MCQ 2Easy RecallPractice
The maximum value of binding energy per nucleon on the curve is approximately
Show answer and why every option is right or wrong
Answer: B. The peak is about 8.8 MeV per nucleon in the iron region, A ≈ 56 (NCERT Class 12 Physics, Chapter 13, page 312).
Why A is wrong: A is wrong because 7.6 MeV per nucleon at A = 238 is the value at the heavy end of the curve, where it has fallen from the peak — it is not the maximum.
Why C is wrong: C is wrong because 931.5 MeV is the energy equivalent of one atomic mass unit, a conversion constant, not a binding energy per nucleon.
Why D is wrong: D is wrong because 28.3 MeV is the TOTAL binding energy of the ⁴He nucleus; per nucleon it is about 7.1 MeV.
MCQ 3Concept TrapPractice
The binding energy per nucleon stays almost constant, near 8.5 MeV, for nuclei from roughly A = 30 to A = 170. This near-constancy is evidence that the nuclear force
Show answer and why every option is right or wrong
Answer: B. If each nucleon interacted only with a few neighbours, total B grows in proportion to A and B/A is constant — which is what the flat region shows (NCERT Class 12 Physics, Chapter 13, page 312).
Why A is wrong: A is wrong because an all-pairs force gives a total binding energy proportional to A(A−1), so B/A would rise steadily with A instead of flattening.
Why C is wrong: C is wrong because the nuclear force is charge-independent and attractive between all nucleon pairs at the relevant separations; nothing in the curve suggests a neutron-repulsive force.
Why D is wrong: D is wrong because a weakening force would make B/A fall throughout the middle range; the observed plateau requires the force to stay the same and merely be short-ranged.
MCQ 4Concept TrapPractice
Beyond A ≈ 170 the binding energy per nucleon decreases slowly with increasing mass number. The principal reason is
Show answer and why every option is right or wrong
Answer: A. Coulomb repulsion is long-range, so every proton repels every other; its contribution grows faster than A and erodes B/A in heavy nuclei (NCERT Class 12 Physics, Chapter 13, page 312 notes that binding energy per nucleon is lower for heavy nuclei; the Coulomb explanation goes beyond NCERT).
Why B is wrong: B is wrong because the nuclear force does not reverse sign with nuclear size; it simply has too short a range to reach across a large nucleus, which is a different statement.
Why C is wrong: C is wrong because heavy nuclei carry proportionally MORE neutrons, not fewer — the neutron excess partly offsets Coulomb repulsion rather than causing the decline.
Why D is wrong: D is wrong because nuclear density is essentially the same for all nuclei; it is not a variable that could explain the fall-off.
MCQ 5Direct ApplicationPractice
From the shape of the binding-energy-per-nucleon curve alone, energy is released when
Show answer and why every option is right or wrong
Answer: B. Energy is released when the products sit higher on the B/A curve than the reactants, and the curve peaks near A ≈ 56, so both ends move toward it (NCERT Class 12 Physics, Chapter 13, page 313).
Why A is wrong: A is wrong because A ≈ 56 is at the peak; splitting it moves the products DOWN the curve, which costs energy rather than releasing it.
Why C is wrong: C is wrong because splitting is favourable only for nuclei heavier than the peak; for light nuclei fragmentation moves products to lower B/A.
Why D is wrong: D is wrong because fusing two nuclei already heavier than the peak produces a product with lower B/A, so that combination absorbs energy.
MCQ 6CalculationPractice
For a nuclide of mass number A = 120 the binding energy per nucleon is 8.50 MeV. The total binding energy of this nucleus is closest to
Show answer and why every option is right or wrong
Answer: C. B = (B/A) × A = 8.50 MeV × 120 = 1.02 × 10³ MeV; the mass number 120 is a counting integer and is exact. The relation B/A = B ÷ A is the definition used in NCERT Class 12 Physics, Chapter 13, page 312.
Why A is wrong: A is wrong because 8.50 × 10² MeV would follow from multiplying by 100 rather than by the stated mass number 120.
Why B is wrong: B is wrong because it comes from dividing 8.50 by 120 — inverting the definition, which already gives energy per nucleon, not per nucleus.
Why D is wrong: D is wrong because 1.28 × 10² MeV is roughly the sum 8.50 + 120, mixing a per-nucleon energy with a dimensionless count.
MCQ 7Direct ApplicationPractice
A student states: "Since the binding energy per nucleon of ²³⁸U (about 7.6 MeV) is less than that of ⁵⁶Fe (about 8.8 MeV), the uranium nucleus has the smaller total binding energy." The statement is
Show answer and why every option is right or wrong
Answer: C. Multiplying each B/A by its mass number gives about 7.6 × 238 ≈ 1.8 × 10³ MeV for uranium and 8.8 × 56 ≈ 4.9 × 10² MeV for iron; the mass numbers are exact counting integers. Total B and B/A rank nuclei differently (NCERT Class 12 Physics, Chapter 13, page 312).
Why A is wrong: A is wrong because the curve's vertical axis is a per-nucleon quantity; a lower point tells you each nucleon is less tightly held, not that the nucleus as a whole is.
Why B is wrong: B is wrong because the two differ by a factor of A, which ranges from 1 to over 200 across the nuclides.
Why D is wrong: D is wrong because the quoted values are the right way round — uranium really does have the lower B/A; the student's error is in the conclusion drawn, not the premise.
MCQ 8CalculationPractice
In the light-nucleus region the curve rises steeply but not smoothly: ⁴He, ¹²C and ¹⁶O each sit noticeably above their immediate neighbours. What do these three nuclides have in common?
Show answer and why every option is right or wrong
Answer: A. ⁴He (2p, 2n), ¹²C (6p, 6n) and ¹⁶O (8p, 8n) are all A = 4n nuclides with N = Z, both even; these are the local maxima superposed on the rising light-nucleus branch (NCERT Class 12 Physics, Chapter 13, page 320).
Why B is wrong: B is wrong because 4 and 16 are not multiples of three at all, and 12 is an even multiple; the shared factor is four, not three.
Why C is wrong: C is wrong because each of the three has exactly equal neutron and proton numbers — neutron excess is a feature of heavy nuclei, not of these light peaks.
Why D is wrong: D is wrong because all three have A well below 56; they lie on the steeply rising branch to the LEFT of the peak.
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Binding Energy Variation: quick recall before you leave
How do you solve a Binding Energy Variation question? A worked example
- 1
Given
Two nuclides from opposite sides of the stability curve:• Nuclide X: A = 20 (exact, counting integer), B/A = 8.00 MeV• Nuclide Y: A = 2.00 × 10² (exact, counting integer), B/A = 7.90 MeV
- 2
Required
Which nuclide has the greater total binding energy, and by what factor — and which is the more tightly bound per nucleon?
- 3
Concept
Total binding energy B and binding energy per nucleon B/A are different rankings of the same nuclei, related by a factor of A. The stability curve's y-axis is B/A; NCERT Class 12 Physics, Chapter 13, page 312, defines it as the average energy needed to remove one nucleon. "More tightly bound" refers to B/A.
- 4
Formula
B = (B/A) × A
- 5
Substitution
B_X = 8.00 MeV × 20
B_Y = 7.90 MeV × 2.00 × 10² - 6
Calculation
B_X = 1.60 × 10² MeV
B_Y = 1.58 × 10³ MeV
Ratio B_Y / B_X = 1.58 × 10³ ÷ 1.60 × 10² = 9.88
The mass numbers 20 and 2.00 × 10² are counting integers and are exact; they do not limit the significant-figure count. The three-significant-figure results follow from the three-significant-figure binding energies alone. - 7
Final answer
Nuclide Y has the greater total binding energy, larger by a factor of 9.88. Nuclide X is the more tightly bound per nucleon (8.00 MeV against 7.90 MeV) and therefore sits higher on the stability curve.
- 8
Common trap
Reading "8.00 MeV versus 7.90 MeV" and concluding X is the more strongly bound nucleus overall. It is not: Y is bound by nearly ten times as much total energy. The curve ranks nuclei by how tightly each individual nucleon is held, which is the quantity that governs whether a nuclear rearrangement releases energy — but it says nothing directly about the total energy locked in a given nucleus.
- 9
Similar NEET-style question
A nuclide of mass number 1.4 × 10² has a total binding energy of 1.19 × 10³ MeV. Determine its binding energy per nucleon and state whether it lies to the left or the right of the peak of the stability curve. *(Answer: 8.50 MeV per nucleon; it lies to the right of the peak, since B/A has fallen below the maximum of about 8.8 MeV while A exceeds 56.)*
What to remember before solving Binding Energy Variation questions
Which Binding Energy Variation formulas do you need for NEET?
1 formula — click to collapse
Mass defect and binding energy
Mass defect = sum of constituent masses minus nuclear mass. Binding energy = mass defect in energy units.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Z | proton number | - |
| N | neutron number | - |
| m_p, m_n | proton, neutron mass | u |
| B | binding energy | MeV |
Valid when
- Non-relativistic
- Stable nucleus
More in Atoms and Nuclei: 3 exam traps and mistakes · 5 formulas · 5 question patterns from its other lessons.
Binding Energy Variation questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 12 Physics Chapter 13, p.313
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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