Physics Class 12 Part II…
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Mass Energy Binding Energy

🎓 Class 12 Physics CBSE Theory Ch 13 – Nuclei ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Mass Energy Binding Energy

આ મૂલ્યાંકન આના પર આધારિત હશે: Mass Energy Binding Energy

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Mass Energy Binding Energy

13.4 Mass–Energy Equivalence (E = mc²)

Before Einstein, mass and energy were thought to be conserved separately. In 1905, Einstein's special relativity changed this for ever: mass is itself a form of energy and the two can be interconverted. The famous relation

E = m c²

says that an object of mass m has a rest-energy E equal to mc², where c = 3 × 10⁸ m s⁻¹ is the speed of light in vacuum. In a nuclear reaction, what we call "energy released" is the difference in total mass-energy between the initial and final states.

Worked Example 13.1 — Energy in 1 g of matter

Calculate the energy equivalent of 1 g of substance.

m = 10⁻³ kg, c = 3 × 10⁸ m/s.

\[ E = mc^2 = 10^{-3} \times (3 \times 10^{8})^2 = 9 \times 10^{13}\ \text{J} \]

That is roughly the energy released by burning 3000 tonnes of coal! In nuclear reactions only a tiny fraction of mass converts to energy, but it is still enormous on the chemical scale.

Energy units in nuclear physics

Because nuclear energies are far larger than chemical energies (eV) but far smaller than macroscopic energies (J), the convenient unit is the MeV (10⁶ eV). The mass-energy of 1 atomic mass unit is

1 u × c² = (1.6605 × 10⁻²⁷ kg)(2.9979 × 10⁸ m/s)² = 1.4924 × 10⁻¹⁰ J = 931.5 MeV

So 1 u ≡ 931.5 MeV/c². This handy conversion lets us go from mass differences in u directly to energies in MeV.

13.4.2 Nuclear Binding Energy & Mass Defect

If a nucleus were just a sum of its protons and neutrons, its mass would equal the total mass of those constituents. But experiment shows a striking surprise: the mass of every stable nucleus is less than the total mass of its individual nucleons.

Take \(^{16}_{8}\mathrm{O}\): it has 8 protons and 8 neutrons.

QuantityValue
Mass of 8 protons8 × 1.00727 u = 8.05816 u
Mass of 8 neutrons8 × 1.00866 u = 8.06928 u
Mass of 8 electrons8 × 0.00055 u = 0.00440 u
Sum of constituents16.13184 u
Atomic mass of \(^{16}\mathrm{O}\) (measured)15.99491 u
Mass defect ΔM0.13691 u ≈ 127.5 MeV/c²
Mass defect: \[ \Delta M = \big[Z\,m_p + (A-Z)\,m_n\big] - M_{\text{nucleus}} \] Binding energy: \[ E_b = \Delta M \cdot c^2 \] This is the energy that must be supplied from outside to dismantle the nucleus into its free constituents — equivalently, the energy that was released when those constituents originally came together.

Worked Example 13.2 — Energy equivalent of 1 u, and BE of ¹⁶O

Find the energy equivalent of 1 u in joules and in MeV. Then express the mass defect of \(^{16}_{8}\mathrm{O}\) in MeV/c².
\[ 1\ \text{u}\cdot c^2 = (1.6605\times 10^{-27})(2.9979\times 10^{8})^2 = 1.4924\times 10^{-10}\ \text{J} \] \[ = \frac{1.4924\times 10^{-10}}{1.602\times 10^{-19}}\ \text{eV} \approx 0.9315 \times 10^{9}\ \text{eV} = 931.5\ \text{MeV} \]

Hence 1 u = 931.5 MeV/c². For \(^{16}\)O, ΔM = 0.13691 u, so

\[ E_b = 0.13691 \times 931.5 = \mathbf{127.5\ MeV} \]

Worked Example 13.3 — Binding energy of nitrogen-14

Obtain the binding energy of \(^{14}_{7}\mathrm{N}\), given m(N-14) = 14.00307 u, m_H = 1.00783 u, m_n = 1.00867 u. (NCERT Exercise 13.1)

Z = 7, A − Z = 7. Using atomic masses (electrons cancel):

\[ \Delta M = 7\,m_H + 7\,m_n - M(^{14}\mathrm{N}) \] \[ = 7(1.00783) + 7(1.00867) - 14.00307 \] \[ = 7.05481 + 7.06069 - 14.00307 = 0.11243\ \text{u} \] \[ E_b = 0.11243 \times 931.5 \approx \mathbf{104.7\ MeV} \] \[ E_{bn} = E_b/A = 104.7 / 14 \approx 7.48\ \text{MeV/nucleon} \]

Binding Energy per Nucleon and the BE/A Curve

A more revealing quantity than the total binding energy is the binding energy per nucleon:

Ebn = Eb / A

Plotting \(E_{bn}\) against the mass number A for all known nuclei produces the famous binding-energy curve shown below — one of the most important graphs in physics.

Mass number A E_bn (MeV/nucleon) 2 4 6 8 10 0 50 100 150 200 250 ²H (1.1) ⁴He (7.07) ¹²C ¹⁶O (7.97) ⁵⁶Fe (8.79 MeV) ¹⁰⁰Mo ²³⁸U (7.6) FUSION → ← FISSION Binding Energy per Nucleon vs A
Fig 13.1: The BE/A curve. Maximum at ⁵⁶Fe (~8.79 MeV/nucleon). Light nuclei gain energy by fusion, heavy nuclei by fission.

Reading the curve — four key conclusions

  • (i) Plateau region: For 30 < A < 170, E_bn is roughly constant at ≈ 8 MeV/nucleon, with a peak of 8.79 MeV at ⁵⁶Fe (A = 56).
  • (ii) Light nuclei (A < 30): Have lower E_bn — they are loosely bound. ²H is just 1.1 MeV/nucleon.
  • (iii) Heavy nuclei (A > 170): Also have lower E_bn — too many protons crammed together suffer Coulomb repulsion. ²³⁸U is only 7.6 MeV/nucleon.
  • (iv) Energy release directions: Going from less-bound to more-bound nuclei releases energy. So fusion of light nuclei (left of peak) and fission of heavy ones (right of peak) both release energy.
Why the plateau? The nuclear force is short-ranged. A nucleon deep inside a large nucleus interacts only with its few immediate neighbours, not with every other nucleon. As nucleons are added, each new one contributes about the same binding energy. This is the saturation property of the nuclear force.

13.5 Nuclear Force — What Holds the Nucleus Together?

For mid-mass nuclei E_bn ≈ 8 MeV — about a million times the binding energy per electron in atoms. To hold protons together against their mutual electrical repulsion, the binding force must be far stronger than the Coulomb force. This is the strong nuclear force.

Key features (summarising decades of scattering experiments 1930–50):

  1. It is much stronger than the Coulomb force at short distances and much, much stronger than gravity.
  2. It has a very short range — falls to essentially zero beyond a few femtometres.
  3. It is charge-independent: n-n, p-p (nuclear part) and p-n forces have the same strength.
  4. At very small separations (≲ 0.8 fm) it becomes strongly repulsive — preventing nucleons from collapsing into each other.
  5. It has no simple closed-form expression like Coulomb's law.
Separation r (fm) U(r) 0 r₀ ≈ 0.8 fm (min) Repulsive (r < r₀) Attractive (r > r₀) 0.4 0.8 1.6 2.4 3.0 Nucleon-pair potential energy
Fig 13.2: U(r) for two nucleons. Strong attraction near 1 fm; sharp repulsive core for r < 0.8 fm; force vanishes beyond a few fm.
Activity 13.2 — Nucleon coin-stack analogy

Stack 5 coins. The ones on top of the stack only "feel" the coins immediately beneath them. Adding a 6th coin doesn't change how strongly the 1st coin feels the 2nd.

How does this analogy explain why E_bn is roughly constant in the mid-A range?
Each nucleon "feels" only its few neighbours within the short range of the strong force, just like each coin feels only the coin above and below. So adding more nucleons does not increase the binding per nucleon — the force has saturated.

Interactive — Binding Energy Calculator

Compute ΔM, E_b and E_bn for any nuclide

Pick an isotope; the calculator computes mass defect, total binding energy and BE per nucleon, and shows where the nuclide sits on the BE/A curve.

Mass defect: 0.137 u Total BE: 127.6 MeV BE/nucleon: 7.97 MeV
A E_bn (MeV) ¹⁶O

Notice how ⁵⁶Fe sits at the peak — it is the most tightly bound nucleus per nucleon.

Competency-Based Questions

Q1 (MCQ). 1 atomic mass unit corresponds to an energy of:

  • (a) 9.31 MeV
  • (b) 93.1 MeV
  • (c) 931.5 MeV
  • (d) 9315 MeV
(c) 1 u × c² = 931.5 MeV.

Q2 (Fill-in-the-blank). The binding energy per nucleon is maximum near A = ____ at a value of approximately ____ MeV/nucleon.

A = 56 (⁵⁶Fe); E_bn ≈ 8.75–8.79 MeV/nucleon.

Q3 (Short Answer). Why is the actual mass of every stable nucleus less than the sum of masses of its constituents?

Because to assemble the nucleus from free nucleons, energy E_b is released (binding energy). By Einstein's E = mc², the released energy corresponds to a loss of mass ΔM = E_b/c² — the mass defect.

Q4 (Numerical). Find the binding energy of \(^{56}_{26}\mathrm{Fe}\) given m(Fe) = 55.934939 u.

ΔM = 26(1.00783) + 30(1.00867) − 55.934939 = 26.2034 + 30.2601 − 55.9349 = 0.5286 u. E_b = 0.5286 × 931.5 ≈ 492.3 MeV. Per nucleon: 492.3/56 ≈ 8.79 MeV/nucleon.

Q5 (HOTS). Why does the BE/A curve drop for very heavy nuclei (A > 170)?

In heavy nuclei, the long-range Coulomb repulsion between many protons (which scales like Z(Z−1)) acts on every pair, while the short-range attractive nuclear force has saturated. The net effect is a reduction in net binding per nucleon, making such nuclei prone to α-decay or fission.

Assertion–Reason Questions

Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: Energy is released when two light nuclei fuse to form a heavier one.

Reason: The BE per nucleon of the heavier product is greater than that of the lighter reactants.

(A) Both correct; the reason explains the assertion. The mass-defect increase manifests as released kinetic energy.

Assertion: The nuclear force between two protons is approximately equal to that between two neutrons.

Reason: The strong nuclear force is charge-independent.

(A) Both correct; the reason explains the assertion.

Assertion: The binding energy per nucleon increases without bound as A increases.

Reason: The strong nuclear force has unlimited range.

(D) Assertion is false (BE/A peaks at A ≈ 56 then decreases) — and reason is also false (the strong force is short-ranged). Wait — both statements are false, so the answer is "both false". Among the standard four options, the closest match is that A is false; R is also false. None of A/B/C/D fits perfectly; the assertion is the wrong-statement candidate. Standard CBSE marking gives (D)-equivalent or "neither correct".

Frequently Asked Questions - Mass Energy Binding Energy

What is the main concept covered in Mass Energy Binding Energy?
In NCERT Class 12 Physics Chapter 13 (Nuclei), "Mass Energy Binding Energy" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Mass Energy Binding Energy useful in real-life applications?
Real-life applications of "Mass Energy Binding Energy" from NCERT Class 12 Physics Chapter 13 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Mass Energy Binding Energy?
Key formulas in "Mass Energy Binding Energy" (NCERT Class 12 Physics Chapter 13 Nuclei) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 13?
NCERT Class 12 Physics Chapter 13 (Nuclei) is structured so each part builds on the previous one. "Mass Energy Binding Energy" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Mass Energy Binding Energy?
CBSE board questions from "Mass Energy Binding Energy" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Mass Energy Binding Energy" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part II – NCERT (2025-26)
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નમસ્તે! 👋 હું ગૌરા છું, Mass Energy Binding Energy માટે તમારું AI ટ્યુટર. આરામથી પાઠ ભણો — જ્યારે પણ કોઈ શંકા થાય, બસ મને પૂછો! હું મદદ માટે અહીં જ છું.

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