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Fission Fusion

🎓 Class 12 Physics CBSE Theory Ch 13 – Nuclei ⏱ ~14 min
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Fission Fusion

13.7 Nuclear Energy — Why Fission and Fusion Both Release Energy

Recall the BE/A curve: the most tightly bound nuclei sit near A ≈ 56 (iron). Lighter nuclei are loosely bound; very heavy nuclei (A > 170) are also less tightly bound because Coulomb repulsion of many protons partly cancels the strong-force attraction. Energy is released whenever loosely-bound nuclei rearrange into more tightly-bound ones.

Two routes to climb up the BE curve:

  • Fission: A heavy nucleus (A ~ 235) splits into two intermediate-mass fragments (each A ~ 100–140). Each fragment is more tightly bound — energy is released.
  • Fusion: Two light nuclei merge into a heavier one with greater BE per nucleon — energy is released.
Energy scale comparison: A typical chemical reaction releases a few eV per atom. A nuclear reaction releases millions of eV (MeV). For the same mass of fuel, nuclear processes liberate roughly a million times more energy than chemical combustion. Burning 1 kg of coal yields ~10⁷ J; fission of 1 kg of ²³⁵U yields ~10¹⁴ J.

13.7.1 Nuclear Fission

Nuclear fission was discovered in 1938 by Otto Hahn and Fritz Strassmann (interpretation by Lise Meitner & Otto Frisch in 1939). When \(^{235}_{92}\mathrm{U}\) absorbs a slow neutron, the resulting \(^{236}\mathrm{U}^{*}\) compound nucleus is so unstable that it splits — typically into two unequal fragments and 2–3 prompt neutrons:

^{1}_{0}n + ^{235}_{92}U → ^{236}_{92}U* → ^{144}_{56}Ba + ^{89}_{36}Kr + 3 ^{1}_{0}n + Q ^{1}_{0}n + ^{235}_{92}U → ^{236}_{92}U* → ^{133}_{51}Sb + ^{99}_{41}Nb + 4 ^{1}_{0}n ^{1}_{0}n + ^{235}_{92}U → ^{140}_{54}Xe + ^{94}_{38}Sr + 2 ^{1}_{0}n

The fragments are radioactive — they sit on the neutron-rich side of the stability valley and undergo successive β-decays until reaching stable nuclei. The Q-value per fission is about 200 MeV, distributed roughly as 165 MeV kinetic energy of fragments + 5 MeV neutron KE + 7 MeV prompt γ-rays + the rest carried away by β/γ from fission products and antineutrinos.

Order-of-magnitude estimate of Q

Take a parent A = 240 nucleus with E_bn ≈ 7.6 MeV splitting into two A = 120 fragments with E_bn ≈ 8.5 MeV.

Gain in BE per nucleon ≈ 8.5 − 7.6 = 0.9 MeV Total Q ≈ 240 × 0.9 ≈ 216 MeV

This rough estimate is in excellent agreement with measurements (~200 MeV).

Neutron-induced fission of ²³⁵U n ²³⁵U Z=92 ²³⁶U* excited ¹⁴⁴Ba Z=56 ⁸⁹Kr Z=36 n n n + ~200 MeV (kinetic + γ)
Fig 13.6: Fission of ²³⁵U by a thermal neutron, producing two daughters and 3 prompt neutrons.

Chain Reaction and Critical Mass

Each fission releases 2–3 neutrons. If at least one of those neutrons triggers another fission, a self-sustaining chain reaction develops. If on average exactly 1 neutron per fission causes the next fission (the multiplication factor k = 1), the reaction is critical and steady — the basis of a power reactor. If k > 1, fissions multiply exponentially — the basis of a fission bomb. The minimum mass of fissile material that allows k ≥ 1 (given a particular geometry) is the critical mass.

Chain Reaction U U U U U U U . . . exponential growth Gen 0 Gen 1 Gen 2 Gen 3
Fig 13.7: A branching chain reaction. Each fission produces >1 daughter neutron, multiplying the rate generation by generation if k > 1.

Nuclear Reactor (Schematic)

A nuclear reactor turns this chain reaction into controlled, steady heat. Five essential components:

  1. Fuel: Usually uranium enriched in ²³⁵U (or ²³⁹Pu). The fissile fuel is shaped into rods.
  2. Moderator: Water, heavy water (D₂O) or graphite. Slows down fast fission neutrons (≈ 1 MeV) to thermal energies (≈ 0.025 eV) where they are most efficiently absorbed by ²³⁵U.
  3. Control rods: Boron or cadmium absorb excess neutrons. Pushing them in slows the reaction; pulling them out speeds it up. They keep k = 1.
  4. Coolant: Water (or liquid metal/gas) carries away the heat.
  5. Shielding: Thick concrete and steel absorb stray neutrons and γ-rays.
Pressurised-water reactor (schematic) REACTOR CORE fuel control rods moderator/coolant: H₂O primary loop (hot) return (cool) Steam Generator Turbine + Generator Condenser concrete shielding
Fig 13.8: Pressurised-water reactor. Fission heats the primary water, a heat-exchanger boils secondary water, steam drives a turbine to make electricity.

Worked Example 13.7 — Energy from 1 kg of ²³⁹Pu

Average energy released per fission of ²³⁹Pu = 180 MeV. How much energy is released if all atoms in 1 kg of ²³⁹Pu undergo fission? (NCERT 13.7)

Number of atoms in 1 kg = (1000 g × N_A) / (239 g/mol) = (1000 × 6.022 × 10²³) / 239 ≈ 2.52 × 10²⁴ atoms.

\[ E_{\text{tot}} = 2.52 \times 10^{24} \times 180\ \text{MeV} = 4.54 \times 10^{26}\ \text{MeV} \]

Converting to joules (1 MeV = 1.6 × 10⁻¹³ J):

\[ E_{\text{tot}} = 4.54 \times 10^{26} \times 1.6 \times 10^{-13} \approx 7.26 \times 10^{13}\ \text{J} \]

Equivalent to burning roughly 2500 tonnes of coal!

13.7.2 Nuclear Fusion — Energy from the Stars

Nuclear fusion is the inverse process: light nuclei combine into a heavier one whose BE per nucleon is greater. Some examples:

^{1}_{1}H + ^{1}_{1}H → ^{2}_{1}H + e⁺ + ν + 0.42 MeV …(p-p) ^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV …(d-d) ^{2}_{1}H + ^{2}_{1}H → ^{3}_{1}H + ^{1}_{1}H + 4.03 MeV …(d-d) ^{2}_{1}H + ^{3}_{1}H → ^{4}_{2}He + n + 17.59 MeV …(d-t)

The Coulomb Barrier

For fusion, the two positively charged nuclei must approach within ~1 fm so the strong force can take over. Their mutual Coulomb repulsion erects a potential barrier of typical height ≈ 400 keV (for two protons). To climb this barrier thermally requires temperatures of order

(3/2) k_B T ≈ 400 keV → T ~ 3 × 10⁹ K

Such high temperatures are why fusion is described as thermonuclear. (The sun's core is "only" 1.5 × 10⁷ K — fusion proceeds there because the high-energy tail of the Maxwell-Boltzmann distribution and quantum tunnelling combine to make protons leak through the barrier.)

Proton-Proton Cycle (energy of the Sun)

In stars like the Sun, hydrogen is "burned" into helium through the p-p cycle:

(i) ¹H + ¹H → ²H + e⁺ + ν + 0.42 MeV (ii) e⁺ + e⁻ → γ + γ + 1.02 MeV (iii) ²H + ¹H → ³He + γ + 5.49 MeV (iv) ³He + ³He → ⁴He + 2 ¹H + 12.86 MeV

Steps (i)-(iii) must occur twice for one (iv). Net effect:

4 ¹H + 2 e⁻ → ⁴He + 2ν + 6γ + 26.7 MeV

Each second, the Sun converts ~6 × 10¹¹ kg of hydrogen into helium, losing ~4 × 10⁹ kg of mass per second to radiation — yet the Sun has enough hydrogen for another ~5 billion years.

Controlled Thermonuclear Fusion

If we can replicate stellar fusion on Earth, we get an essentially unlimited, low-radioactive-waste energy source. The challenge is plasma confinement — heating fuel to 10⁸ K without a container melting. Two main approaches:

  • Magnetic confinement (tokamak): toroidal magnetic fields confine the plasma. Examples: ITER (international project), India's ADITYA and SST-1.
  • Inertial confinement: high-power lasers (or ion beams) compress and heat a tiny fuel pellet (e.g. NIF, USA).

Worked Example 13.8 — Lamp powered by deuterium fusion

For how long can a 100 W electric lamp be kept glowing by fusing 2.0 kg of deuterium via ²H + ²H → ³He + n + 3.27 MeV? (NCERT 13.8)

Number of D atoms in 2.0 kg: (2000 × 6.022 × 10²³)/2 = 6.022 × 10²⁶.

Two deuterons fuse per reaction → number of reactions = 6.022 × 10²⁶ / 2 = 3.011 × 10²⁶.

Total energy = 3.011 × 10²⁶ × 3.27 MeV × 1.6 × 10⁻¹³ J/MeV ≈ 1.576 × 10¹⁴ J.

\[ t = \frac{E}{P} = \frac{1.576 \times 10^{14}}{100} = 1.576 \times 10^{12}\ \text{s} \approx \mathbf{5 \times 10^{4}\ \text{years}} \]

Worked Example 13.9 — Coulomb barrier height

Calculate the height of the Coulomb potential barrier for a head-on collision of two deuterons, treating each as a hard sphere of radius 2.0 fm. (NCERT 13.9)

At touch, separation = 2 × 2.0 fm = 4.0 × 10⁻¹⁵ m.

\[ U = \frac{1}{4\pi\varepsilon_0}\frac{e\cdot e}{r} = \frac{(9 \times 10^{9})(1.6 \times 10^{-19})^2}{4 \times 10^{-15}} \] \[ = \frac{2.304 \times 10^{-28}}{4 \times 10^{-15}} = 5.76 \times 10^{-14}\ \text{J} \]

Converting to MeV: U ≈ 5.76 × 10⁻¹⁴ / 1.6 × 10⁻¹³ = 0.36 MeV ≈ 360 keV.

Interactive — Mass-to-Energy Calculator

Q-value calculator

Pick a nuclear reaction; see the mass defect and Q-value released.

Mass before: 5.030 u Mass after: 5.011 u ΔM: 0.019 u Q: 17.59 MeV
BEFORE ²H + ³H 5.030 u AFTER ⁴He + n 5.011 u + Q 17.59 MeV (=ΔM·c²)

Competency-Based Questions

Q1 (MCQ). The energy released per fission of ²³⁵U is approximately:

  • (a) 2 MeV
  • (b) 20 MeV
  • (c) 200 MeV
  • (d) 2000 MeV
(c) About 200 MeV per fission.

Q2 (MCQ). Which of the following is NOT a moderator in a thermal reactor?

  • (a) Heavy water
  • (b) Graphite
  • (c) Cadmium
  • (d) Light water
(c) Cadmium is a strong neutron absorber; it is used in control rods, not as a moderator.

Q3 (Short Answer). Why does fusion require very high temperatures?

Both nuclei are positively charged and electrically repel each other. They must approach within ≈ 1 fm so the short-range strong force can bind them. Climbing this Coulomb barrier (~ 0.4 MeV for two protons) needs kinetic energies attainable only at thermonuclear temperatures (~ 10⁷ K or higher).

Q4 (Numerical). Estimate the mass converted to energy when ²³⁵U undergoes fission releasing 200 MeV.

Δm = E/c² = 200 MeV / (931.5 MeV/u) ≈ 0.215 u ≈ 3.57 × 10⁻²⁸ kg per fission. About 0.1% of the original nuclear mass.

Q5 (HOTS). Why is fusion considered cleaner than fission as an energy source?

(i) Fusion fuel (deuterium) is abundant in seawater. (ii) Primary fusion products (e.g. ⁴He) are stable and non-radioactive — unlike fission products which are highly radioactive isotopes with long half-lives. (iii) Fusion produces no long-lived radioactive waste, no transuranic elements, and cannot run away (a malfunctioning fusion reactor simply stops). (iv) No risk of nuclear-weapon proliferation from fuel.

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: A nuclear reactor uses moderators like heavy water to slow down neutrons.

Reason: Slow (thermal) neutrons are more efficiently captured by ²³⁵U to cause fission.

(A) Both correct, and the reason explains the assertion.

Assertion: Energy is released in both fission of ²³⁵U and fusion of deuterium.

Reason: In each case, the products have higher binding energy per nucleon than the reactants.

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

Assertion: Critical mass of fissile material depends on its geometry and surroundings.

Reason: Neutron leakage is determined by surface-area-to-volume ratio.

(A) Both correct; the reason explains the assertion. A sphere has the smallest S:V ratio and therefore the lowest critical mass.

Frequently Asked Questions - Fission Fusion

What is the main concept covered in Fission Fusion?
In NCERT Class 12 Physics Chapter 13 (Nuclei), "Fission Fusion" 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 Fission Fusion useful in real-life applications?
Real-life applications of "Fission Fusion" 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 Fission Fusion?
Key formulas in "Fission Fusion" (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. "Fission Fusion" 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 Fission Fusion?
CBSE board questions from "Fission Fusion" 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 "Fission Fusion" 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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