Nuclear Energy & Fission

A Million Times Chemistry

The binding-energy curve's long flat middle (A = 30 to 170, ~8.0 MeV) with drooping ends is an energy map: moving nucleons from loosely bound arrangements to tightly bound ones releases the difference. Greater binding = less total mass — so transformations toward tighter binding shed mass as energy.

Two routes exist:

• Fission: a heavy nucleus (A > 170 region, ~7.6 MeV/nucleon) splits into middle-mass fragments (~8.5 MeV/nucleon).
• Fusion: light nuclei (A < 30) merge into a heavier, tighter nucleus.

The scale of the prize

Chemical reactions (coal, petroleum) turn over eV per atom; nuclear reactions turn over MeV per nucleus — a factor of a million:

Fuel (1 kg)	Energy
Coal (burning)	~$10^{7}$ J
Uranium (fission)	~$10^{14}$ J

Fission: Splitting the Heavyweights

Beyond natural decays, nuclei can be bombarded with protons, neutrons, alphas… The most important neutron-induced reaction is fission. A slow neutron absorbed by $^{235}_{92}$U forms $^{236}_{92}$U, which promptly splits:

$$^{1}_{0}n + {}^{235}_{92}U \to {}^{236}_{92}U \to {}^{144}_{56}Ba + {}^{89}_{36}Kr + 3\,^{1}_{0}n$$

The same capture can yield other fragment pairs:

$$^{1}_{0}n + {}^{235}_{92}U \to {}^{133}_{51}Sb + {}^{99}_{41}Nb + 4\,^{1}_{0}n \qquad\text{or}\qquad {}^{140}_{54}Xe + {}^{94}_{38}Sr + 2\,^{1}_{0}n$$

Note the bookkeeping in each: A and Z balance on both sides, and 2-4 fresh neutrons emerge — the seed of chain reactions.

The fragments are radioactive

Fragment nuclei inherit the parent's neutron-rich N/Z ratio — too high for their mass range — so they emit beta particles in succession until they reach stable end products.

The 200 MeV estimate

From the curve (as computed in Section 3): A = 240 at 7.6 MeV/nucleon → two A = 120 fragments at 8.5 MeV/nucleon; gain ≈ 0.9 MeV × 240 ≈ 216 MeV. The actual Q value ≈ 200 MeV per fission, appearing first as kinetic energy of fragments and neutrons, then degraded to heat in the surroundings.

• Controlled fission chain → nuclear reactors (the heat drives turbines).
• Uncontrolled chain → the atom bomb.

[NEET Important] Q value definition (NCERT summary): Q = final kinetic energy - initial kinetic energy = (sum of initial masses - sum of final masses)$c^2$. Exothermic: Q > 0 (products lighter); endothermic: Q < 0 (energy must be supplied).

How Nuclear Equations 'Balance' (NCERT Example 13.4's Wisdom)

Three subtle questions, answered NCERT's way:

(a) Are nuclear equations balanced like chemical ones?

A chemical equation balances atoms of each element (atoms are merely regrouped). In nuclear reactions elements transmute — atom counts per element are NOT conserved. What balances instead: the number of protons and the number of neutrons, each separately (at ordinary energies; strictly, total charge and baryon number).

(b) If proton and neutron numbers are conserved, where does the mass-energy conversion happen?

The rest masses of free protons and neutrons match on both sides — but binding energy contributes (negatively) to each nucleus's mass, and the total binding energy differs between the two sides. That binding-energy difference is what appears as released (or absorbed) energy. 'The difference in the total mass of nuclei on the two sides gets converted into energy.'

(c) Is mass-energy interconversion exclusive to nuclear reactions?

No — chemical reactions do it too, in principle: chemical binding energy also gives a (tiny) negative mass contribution, so reacting molecules' total mass changes. But chemical mass defects are ~a million times smaller, hence undetectable by weighing — the source of the (incorrect) impression that only nuclear reactions convert mass.

Key Point: Energy release in ANY reaction = binding-energy gain = mass loss. Nuclear reactions are only quantitatively, not qualitatively, special.

[JEE Tip] Q-value machinery for numericals: $Q = [\Sigma m_{initial} - \Sigma m_{final}] \times 931.5$ MeV with masses in u. Atomic masses may be used throughout when electron counts balance (they do in fission and most reactions; beta decays need care).

Solved Examples

Example 1: Energy from 1 kg of plutonium (NCERT Exercise 13.7)

The average fission of $^{239}_{94}$Pu releases 180 MeV. How much energy is released if all atoms in 1 kg of pure $^{239}$Pu fission?

Solution:

1. Number of atoms: $N = \dfrac{1000}{239} \times 6.023 \times 10^{23} \approx 2.52 \times 10^{24}$.
2. Total energy: $E = 2.52 \times 10^{24} \times 180$ MeV $= 4.53 \times 10^{26}$ MeV.
3. In joules: $\times 1.6 \times 10^{-13}$ → $E \approx 7.3 \times 10^{13}$ J.
4. Takeaway: ~$10^{14}$ J per kilogram — NCERT's million-times-coal claim, verified by direct count.

Example 2: Is iron fission possible? (NCERT Exercise 13.6)

Could $^{56}_{26}$Fe (55.93494 u) split into two $^{28}_{13}$Al (27.98191 u)? Compute Q.

Solution:

1. Q value: $Q = [m(^{56}\text{Fe}) - 2m(^{28}\text{Al})] \times 931.5$.
2. Compute: $[55.93494 - 55.96382] \times 931.5 = (-0.02888)(931.5) \approx -26.9$ MeV.
3. Verdict: Q < 0 — energetically impossible spontaneously; ~27 MeV would have to be SUPPLIED.
4. Takeaway: iron sits at the curve's peak — nothing pays from there. Fission profits only for heavy nuclei on the drooping right flank.

Example 3: Q values of two reactions (NCERT Exercise 13.5)

Find Q for (i) $^{1}_{1}H + {}^{3}_{1}H \to {}^{2}_{1}H + {}^{2}_{1}H$ and (ii) $^{12}_{6}C + {}^{12}_{6}C \to {}^{20}_{10}Ne + {}^{4}_{2}He$, given m($^2$H) = 2.014102 u, m($^3$H) = 3.016049 u, m($^{12}$C) = 12.000000 u, m($^{20}$Ne) = 19.992439 u, m($^1$H) = 1.007825 u, m($^4$He) = 4.002603 u.

Solution:

1. (i) $Q = [m(^1H) + m(^3H) - 2m(^2H)] \times 931.5 = [1.007825 + 3.016049 - 4.028204] \times 931.5$.
2. $= (-0.004330)(931.5) \approx -4.03$ MeV → endothermic (needs energy).
3. (ii) $Q = [2(12.000000) - 19.992439 - 4.002603] \times 931.5 = (0.004958)(931.5)$.
4. $\approx +4.62$ MeV → exothermic (releases energy).
5. Takeaway: the sign of the mass difference IS the verdict; carbon-carbon fusion powers heavy stars, while reaction (i) runs backward for free.

Example 4: Balancing a fission equation [Board Numerical]

Complete: $^{1}_{0}n + {}^{235}_{92}U \to {}^{140}_{54}Xe + {}^{?}_{?}X + 2\,^{1}_{0}n$. Identify X.

Solution:

1. Mass numbers: 1 + 235 = 140 + A + 2 → A = 94.
2. Atomic numbers: 0 + 92 = 54 + Z + 0 → Z = 38 (strontium).
3. Answer: $^{94}_{38}$Sr — NCERT Eq. 13.12's fragment pair.
4. Takeaway: protons and neutrons balance separately — two quick sums identify any missing fragment.

Example 5: Energy per fission from curve values [Board Numerical]

Estimate the energy released when $^{236}$U (treat as A = 236 at 7.6 MeV/nucleon) splits into fragments averaging 8.5 MeV/nucleon.

Solution:

1. Before: 236 × 7.6 ≈ 1794 MeV of binding.
2. After: 236 × 8.5 ≈ 2006 MeV.
3. Released: ≈ 212 MeV — the canonical '~200 MeV per fission'.
4. Takeaway: you never need fragment identities for the estimate; the curve alone prices the split.

Example 6: Why are fission fragments beta-active? [NEET Conceptual]

Explain why fission fragments are radioactive and what decay they undergo.

Solution:

1. Heavy nuclei carry N/Z ≈ 1.55 (e.g. U-235); stable middle-mass nuclei want N/Z ≈ 1.2-1.3.
2. Fragments inherit the parent's high ratio → they are neutron-rich for their size.
3. They therefore emit beta-minus particles in succession (n → p conversions) until reaching stable end products — NCERT's exact statement.
4. Takeaway: fission's radioactivity problem (waste) is the N/Z mismatch playing out.

Example 7: Counting fissions for a power station [JEE Numerical]

A reactor delivers 1000 MW of fission heat. How many $^{235}$U fissions occur per second (200 MeV each)?

Solution:

1. Per fission: 200 MeV = $200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}$ J.
2. Rate: $\dfrac{10^9}{3.2 \times 10^{-11}} \approx 3.1 \times 10^{19}$ fissions per second.
3. Takeaway: $10^{19}$-scale event rates from gram-scale fuel consumption — divide power by energy-per-event, always.

Example 8: Chemical vs nuclear mass defect [Conceptual Numerical]

Burning 1 kg of coal releases $10^7$ J. What mass does the products' total fall by, and why is it unmeasurable?

Solution:

1. Mass loss: $\Delta m = \dfrac{E}{c^2} = \dfrac{10^7}{9 \times 10^{16}} \approx 1.1 \times 10^{-10}$ kg — a tenth of a microgram.
2. Fractionally: ~$10^{-13}$ of the fuel mass — hopeless to weigh.
3. Contrast: fission converts ~$10^{-3}$ of the mass (0.09%) — a billion times more, easily seen in mass spectrometry.
4. Takeaway: NCERT Example 13.4(c): chemistry converts mass too, just a million times less per event — the 'only nuclear reactions convert mass' impression is strictly wrong.

Example 9: What is conserved? (NCERT Example 13.4a)

In what sense is $^{1}_{0}n + {}^{235}_{92}U \to {}^{144}_{56}Ba + {}^{89}_{36}Kr + 3\,^{1}_{0}n$ balanced?

Solution:

1. Not in atoms of each element — barium and krypton did not exist beforehand (transmutation!).
2. Balanced in: proton number (92 = 56 + 36) and neutron number (1 + 143 = 88 + 53 + 3 → 144 = 144 ✔), each separately.
3. (At very high energies, the strict laws are conservation of total charge and baryon number.)
4. Takeaway: chemical balancing preserves atoms; nuclear balancing preserves nucleon types.

Example 10: Uranium vs coal, per kilogram [Board Numerical]

Verify the million-fold claim: compare energy per kilogram of U-235 fission (200 MeV/nucleus) vs coal ($10^7$ J/kg).

Solution:

1. Uranium: $N = \dfrac{1000}{235} \times 6.023 \times 10^{23} = 2.56 \times 10^{24}$ nuclei/kg.
2. Energy: $2.56 \times 10^{24} \times 3.2 \times 10^{-11}$ J ≈ $8.2 \times 10^{13}$ J ≈ $10^{14}$ J.
3. Ratio: $\dfrac{10^{14}}{10^{7}} = 10^{7}$ — ten million times coal (NCERT's 'million times' is the conservative order).
4. Takeaway: the eV-vs-MeV gap per event, times similar atom counts, gives the macroscopic million-plus factor.