Mass-Energy Equivalence & Nuclear Binding Energy

Einstein's Bombshell: Mass IS Energy

Before special relativity, mass and energy were conserved separately. Einstein showed that mass is another form of energy — convertible into kinetic energy and vice versa — through the most famous equation in physics:

$$E = mc^2$$

with c ≈ $3 \times 10^8$ m/s. The exchange rate is staggering: 1 gram of matter ↔ $9 \times 10^{13}$ J (NCERT Example 13.2) — roughly a large power station's daily output.

In any reaction, the conservation law now reads: initial energy = final energy, provided the energy locked in mass is included. Nuclear physics is where this principle shows itself most convincingly.

The golden conversion factor

Converting 1 u to energy (NCERT Example 13.3):

$$1\text{ u} \times c^2 = 1.6605 \times 10^{-27} \times (2.9979 \times 10^8)^2 = 1.4924 \times 10^{-10} \text{ J}$$

$$\boxed{1\text{ u} = 931.5 \text{ MeV}/c^2}$$

Every mass-defect calculation in this chapter runs through this factor. Also handy: 1 MeV = $1.6 \times 10^{-13}$ J.

Mass Defect and Binding Energy

Here's the chapter's central surprise: a nucleus weighs less than its parts.

The oxygen-16 audit (NCERT's worked case)

Expected mass of $^{16}_{8}$O nucleus (8 protons + 8 neutrons):

• 8 neutrons: 8 × 1.00866 u; 8 protons: 8 × 1.00727 u → total 16.12744 u (using 8 × 2.01593 u).
• Measured: atomic mass 15.99493 u minus 8 electrons (8 × 0.00055 u) = 15.99053 u.
• Shortfall: 0.13691 u.

This difference is the mass defect:

$$\Delta M = [Zm_p + (A - Z)m_n] - M$$

What the missing mass means

By $E = mc^2$, the bound nucleus has less energy than its separated constituents. To pull the nucleus apart into free nucleons you must supply $\Delta Mc^2$; conversely, assembling a nucleus from free nucleons releases that energy. This is the binding energy:

$$E_b = \Delta Mc^2$$

For oxygen-16: $E_b = 0.13691 \times 931.5 = 127.5$ MeV.

Binding energy per nucleon

The more useful measure of how tightly a nucleus is glued:

$$E_{bn} = \frac{E_b}{A}$$

— the average energy needed to remove one nucleon. For oxygen-16: 127.5/16 ≈ 8.0 MeV per nucleon.

[JEE Tip] Bookkeeping discipline: when given atomic masses, either subtract Z electron masses to get nuclear masses, or use hydrogen-ATOM masses ($m_H$ = 1.007825 u) for protons — the electrons then cancel: $\Delta M = [Zm_H + (A-Z)m_n] - M_{atom}$. NCERT's exercise data supplies $m_H$ for exactly this reason.

The Binding-Energy Curve: Physics' Most Consequential Graph

Plot $E_{bn}$ against mass number A for all nuclei (NCERT Fig. 13.1). Its anatomy:

1. A broad flat plateau: $E_{bn}$ ≈ constant (~8.0 MeV) for 30 < A < 170 — practically independent of A.
2. The maximum: about 8.75 MeV at A = 56 (iron's neighbourhood — the most tightly bound matter in the universe).
3. Droop at the heavy end: down to 7.6 MeV at A = 238.
4. Droop at the light end: $E_{bn}$ lower for A < 30 (with local spikes at $^4$He, $^{16}$O — evidence of shell structure, per Points to Ponder).

Reading the curve: four conclusions

• (i) The nuclear force is attractive and strong enough to yield ~MeV-per-nucleon binding.
• (ii) The plateau's constancy ⇒ the force is short-ranged and saturates: a nucleon inside a large nucleus feels only its p nearest neighbours, so its binding ≈ pk regardless of A. Adding distant nucleons changes nothing for it.
• (iii) Fission pays: A = 240 sits at ~7.6 MeV; two A = 120 fragments sit at ~8.5 MeV. Breaking the heavy nucleus tightens the binding — energy is released (~0.9 MeV × 240 ≈ 216 MeV). The germ of nuclear reactors.
• (iv) Fusion pays too: joining two very light nuclei (A ≤ 10) yields a heavier nucleus with higher $E_{bn}$ — again energy released. The Sun's engine.

Key Point: Greater binding energy ⇒ smaller total mass. Any transformation moving nuclei up the curve (toward iron, from either side) releases energy. Both fission and fusion are just downhill rolls toward A ≈ 56.

[NEET Important] The four numbers — 8.0 MeV plateau, 8.75 MeV max at A = 56, 7.6 MeV at A = 238, range 30 < A < 170 — are quizzed verbatim. So is 'why is the curve flat?' (short range + saturation).

Solved Examples

Example 1: Energy in a gram (NCERT Example 13.2)

Calculate the energy equivalent of 1 g of substance.

Solution:

1. Formula: $E = mc^2 = 10^{-3} \times (3 \times 10^8)^2$.
2. Compute: $E = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}$ J.
3. Takeaway: total conversion of one gram powers a city for a day — the reason nuclear energetics dwarfs chemistry.

Example 2: 1 u in MeV (NCERT Example 13.3)

Find the energy equivalent of one atomic mass unit in joules and MeV; express oxygen-16's mass defect in MeV/$c^2$.

Solution:

1. Joules: $E = 1.6605 \times 10^{-27} \times (2.9979 \times 10^8)^2 = 1.4924 \times 10^{-10}$ J.
2. eV: $\dfrac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} = 0.9315 \times 10^9$ eV = 931.5 MeV.
3. Oxygen: $\Delta M = 0.13691$ u = $0.13691 \times 931.5 = 127.5$ MeV/$c^2$.
4. Takeaway: 931.5 MeV per u — the chapter's master conversion; oxygen-16 needs 127.5 MeV to fully dismantle.

Example 3: Binding energy of nitrogen-14 (NCERT Exercise 13.1)

Given $m(^{14}_{7}\text{N})$ = 14.00307 u (atomic), $m_H$ = 1.007825 u, $m_n$ = 1.008665 u, find $E_b$.

Solution:

1. Atomic-mass bookkeeping: $\Delta M = [7m_H + 7m_n] - m(^{14}\text{N})$.
2. Compute constituents: $7 \times 1.007825 + 7 \times 1.008665 = 7.054775 + 7.060655 = 14.11543$ u.
3. Defect: $\Delta M = 14.11543 - 14.00307 = 0.11236$ u.
4. Energy: $E_b = 0.11236 \times 931.5 \approx 104.7$ MeV.
5. Takeaway: ≈ 7.5 MeV per nucleon — nitrogen sits below the plateau, as a light nucleus should.

Example 4: Iron vs bismuth (NCERT Exercise 13.2)

Find the binding energies of $^{56}_{26}$Fe (55.934939 u) and $^{209}_{83}$Bi (208.980388 u), and compare $E_{bn}$.

Solution:

1. Iron: $\Delta M = [26 \times 1.007825 + 30 \times 1.008665] - 55.934939 = [26.20345 + 26.25995] - 55.934939 = 0.528461$ u.
2. $E_b = 0.528461 \times 931.5 \approx 492.3$ MeV; $E_{bn} = 492.3/56 \approx 8.79$ MeV.
3. Bismuth: $\Delta M = [83 \times 1.007825 + 126 \times 1.008665] - 208.980388 = [83.649475 + 127.09179] - 208.980388 = 1.760877$ u.
4. $E_b = 1.760877 \times 931.5 \approx 1640.3$ MeV; $E_{bn} = 1640.3/209 \approx 7.85$ MeV.
5. Takeaway: iron out-binds bismuth per nucleon (8.79 vs 7.85 MeV) — the curve's peak versus its heavy-side droop, in real numbers.

Example 5: Tearing apart a copper coin (NCERT Exercise 13.3)

A 3.0 g coin of pure $^{63}_{29}$Cu (62.92960 u): how much energy separates all its neutrons and protons?

Solution:

1. Atoms in the coin: $N = \dfrac{3.0}{62.92960} \times 6.023 \times 10^{23} \approx 2.871 \times 10^{22}$.
2. Per-nucleus defect: $\Delta M = [29 \times 1.007825 + 34 \times 1.008665] - 62.92960 = [29.226925 + 34.29461] - 62.92960 = 0.591935$ u.
3. Per-nucleus energy: $0.591935 \times 931.5 \approx 551.4$ MeV.
4. Total: $E = 2.871 \times 10^{22} \times 551.4$ MeV $\approx 1.58 \times 10^{25}$ MeV $= 2.5 \times 10^{12}$ J.
5. Takeaway: dismantling one small coin nucleus-by-nucleus costs the energy of a small power plant running for hours — binding energies are colossal at everyday scales.

Example 6: Reading the curve for fission [Board Conceptual]

Using curve values, estimate the energy released when an A = 240 nucleus splits into two A = 120 fragments.

Solution:

1. Before: $E_{bn} \approx 7.6$ MeV → total binding ≈ 240 × 7.6 = 1824 MeV.
2. After: $E_{bn} \approx 8.5$ MeV → total ≈ 240 × 8.5 = 2040 MeV.
3. Gain: ≈ 0.9 MeV per nucleon × 240 ≈ 216 MeV released — NCERT's own estimate of ~200 MeV per fission.
4. Takeaway: tighter binding after = energy out; the curve is a lookup table for reaction energetics.

Example 7: Why the plateau? [Board Derivation]

Explain the constancy of $E_{bn}$ for 30 < A < 170 from the nuclear force's short range.

Solution:

1. A nucleon deep inside a large nucleus interacts only with neighbours within the force's few-fm range — say a maximum of p of them.
2. Its binding energy ≈ pk (k an energy constant), independent of how big the nucleus is beyond that.
3. Most nucleons of a large nucleus are interior ones, so the average — $E_{bn}$ — stays ≈ pk as A grows: the plateau.
4. This 'a nucleon only feels its neighbours' property is the saturation of the nuclear force. ∎

Example 8: Mass of the products [JEE Conceptual]

In an exothermic nuclear reaction, how does the total rest mass of products compare with reactants?

Solution:

1. Energy released means final kinetic energy > initial — that energy comes from rest mass.
2. Products are lighter: $\Delta m = Q/c^2$.
3. Equivalently: products are more tightly bound (more binding energy = less mass).
4. Takeaway: 'released energy' ↔ 'lost mass' ↔ 'gained binding' — three phrasings of one fact; exams rotate among them.

Example 9: Binding energy per nucleon of helium-4 [NEET Numerical]

Given $m(^4_2\text{He})$ = 4.002603 u (atomic), $m_H$ = 1.007825 u, $m_n$ = 1.008665 u, find $E_{bn}$.

Solution:

1. Defect: $\Delta M = [2 \times 1.007825 + 2 \times 1.008665] - 4.002603 = 4.03298 - 4.002603 = 0.030377$ u.
2. Binding energy: $E_b = 0.030377 \times 931.5 \approx 28.3$ MeV.
3. Per nucleon: $E_{bn} = 28.3/4 \approx 7.07$ MeV.
4. Takeaway: helium-4's unusually high $E_{bn}$ for its size (a spike on the curve) signals shell structure — and explains why alpha particles are nature's favourite ejected package.

Example 10: How much mass does binding steal? [JEE Numerical]

What fraction of oxygen-16's constituent mass is 'missing' as binding energy?

Solution:

1. Fraction: $\dfrac{\Delta M}{M_{constituents}} = \dfrac{0.13691}{16.12744}$.
2. Compute: ≈ 0.0085 → about 0.85%.
3. Takeaway: nuclear binding shaves off nearly 1% of the mass — chemistry's binding shaves ~$10^{-9}$. The million-fold energy gap between nuclear and chemical reactions in one ratio.