Nuclear Energy - From Splitting Atoms to Bottling a Star - eduPhysics Nuclear Fission and Fusion: Notes, MCQs, PYQs, Numericals

• CLASS 12 • JEE • NEET • BOARDS

1. The Big Idea: Where Does Nuclear Energy Come From?

A nucleus weighs less than the sum of its separate protons and neutrons. This missing mass is the mass defect, and Einstein’s relation turns it into the binding energy, the energy you would have to supply to pull the nucleus apart.

The quantity that matters is binding energy per nucleon (B/A), which measures how tightly each nucleon is held. Plot it against mass number A and you get the single most important graph in nuclear physics.

Nucleus	Mass number A	B/A (MeV per nucleon)	Comment
²H (deuteron)	2	≈ 1.11	Very loosely bound
⁴He	4	≈ 7.07	Unusually stable “alpha” peak
¹²C	12	≈ 7.68	Product of triple-alpha process
⁵⁶Fe	56	≈ 8.79	Peak of the curve (with ⁶²Ni)
¹¹⁸Sn (typical fragment)	~118	≈ 8.5	Fission fragments sit here
²³⁵U	235	≈ 7.6	Heavy, less tightly bound

2. Nuclear Fission: Splitting the Heavy

Fission is the splitting of a heavy nucleus into two (rarely three) lighter nuclei, accompanied by neutrons and a large release of energy. It was discovered in 1938 by Otto Hahn and Fritz Strassmann, and explained theoretically by Lise Meitner and Otto Frisch, who also coined the word “fission”, borrowed from biology.

A slow (thermal) neutron strikes ²³⁵U, which briefly becomes an excited ²³⁶U nucleus that wobbles like a liquid drop and splits:

n + ²³⁵U → ²³⁶U* → ¹⁴¹Ba + ⁹²Kr + 3n + ≈ 200 MeV

There are many possible fragment pairs (e.g. Xe–Sr, Ba–Kr). On average, 2.5 neutrons are emitted per fission, and this surplus is what makes a chain reaction possible.

Where do the 200 MeV go?

Form of energy	Approx. share (MeV)
Kinetic energy of fission fragments (becomes heat)	≈ 168
Kinetic energy of prompt neutrons	≈ 5
Prompt gamma rays	≈ 7
Beta particles, delayed gammas, antineutrinos (decay of fragments)	≈ 20
Total	≈ 200

Quick sanity check: B/A of ²³⁵U ≈ 7.6 MeV, B/A of the fragments ≈ 8.5 MeV. Gain ≈ 0.9 MeV × 235 nucleons ≈ 210 MeV, consistent with the ≈ 200 MeV figure.

The chain reaction

Each fission releases neutrons that can trigger further fissions. The multiplication factor k is the ratio of neutrons in one generation to the previous one:

k = (neutrons in a generation) / (neutrons in the previous generation)

Value of k	State	Consequence
k < 1	Subcritical	Chain reaction dies out
k = 1	Critical	Steady, controlled power: a nuclear reactor
k > 1	Supercritical	Exponential growth: an atomic bomb if uncontrolled

Critical mass: the minimum mass of fissile material for which k ≥ 1. Below it, too many neutrons leak out through the surface.
Why slow neutrons? Thermal neutrons (≈ 0.025 eV) are captured far more readily by ²³⁵U than fast neutrons (≈ 2 MeV), so reactors slow neutrons down with a moderator.
Delayed neutrons: about 0.65% of fission neutrons arrive seconds late. This tiny fraction slows the reactor’s response from microseconds to seconds, which is what makes human control possible.
Natural uranium contains only 0.7% ²³⁵U (99.3% is ²³⁸U). Power reactors use natural or slightly enriched (3–5%) fuel, while weapons need > 90%.

Anatomy of a nuclear reactor

Component	Purpose	Typical materials
Fuel	Fissile material that sustains the chain reaction	Natural / enriched UO₂, MOX
Moderator	Slows fast neutrons to thermal energies through elastic collisions	Heavy water (D₂O), graphite, light water
Control rods	Absorb excess neutrons to hold k = 1; inserted fully to shut down	Cadmium, boron, hafnium
Coolant	Carries heat from the core to the steam generator	Water, heavy water, liquid sodium, CO₂
Shielding / containment	Stops neutrons and gamma rays and contains any release	Thick concrete, steel, lead

Why a good moderator is made of light nuclei: in an elastic collision a neutron loses the most energy when it hits a particle of nearly equal mass. Heavy water absorbs very few neutrons, which is why India’s PHWRs can run on natural uranium.

Breeder reactors and fertile materials

Some isotopes are not fissile themselves but can be converted into fissile fuel by neutron capture. These are called fertile materials:

²³⁸U + n → ²³⁹U → ²³⁹Np → ²³⁹Pu (fissile)

²³²Th + n → ²³³Th → ²³³Pa → ²³³U (fissile)

3. Nuclear Fusion: Joining the Light

Fusion is the merging of two light nuclei into a heavier nucleus. Because the product sits higher on the binding-energy curve, the difference is released as energy. Per unit mass, fusion releases about three to four times the energy of fission.

²H + ³H → ⁴He + n + 17.6 MeV (D–T, the easiest)

²H + ²H → ³He + n + 3.27 MeV

²H + ²H → ³H + ¹H + 4.03 MeV

²H + ³He → ⁴He + ¹H + 18.3 MeV

The Coulomb barrier problem

Nuclei are positively charged and repel each other. To fuse, they must approach within about 10⁻¹⁵ m, where the short-range strong force finally takes over. For two deuterons the barrier is roughly 0.4–0.5 MeV. Heating matter until particles collide with such energy means temperatures of 10⁸ K or more (a state of matter called plasma), which is why fusion is also called thermonuclear fusion.

Fission vs Fusion at a glance

Feature	Fission	Fusion
Process	Heavy nucleus splits	Light nuclei combine
Typical fuel	²³⁵U, ²³⁹Pu	²H, ³H (³He)
Energy per reaction	≈ 200 MeV	17.6 MeV (D–T)
Energy per kg of fuel	≈ 8 × 10¹³ J	≈ 3.4 × 10¹⁴ J (D–T)
Conditions	Slow neutrons, critical mass	~10⁸ K, high density, confinement
Chain reaction	Yes (neutron-driven)	No (needs sustained heating)
Waste	Long-lived radioactive	Mostly short-lived activated materials; helium is the main product
Status	Commercial for 70+ years	Experimental; not yet net-electric
Natural example	Not in nature (except Oklo reactors)	Sun and all stars

4. Energy Generation in Stars

A star is a battle between two forces. Gravity tries to crush it; thermal pressure from fusion in the core pushes outward. A star shines for as long as this hydrostatic equilibrium can be maintained by fuel burning. When one fuel runs out, the core contracts, heats up and ignites the next one.

Hydrogen burning I: the proton–proton (p–p) chain

This is the main engine of the Sun and other stars of about one solar mass or less, at a core temperature of about 1.5 × 10⁷ K.

Step	Reaction	Notes
1	¹H + ¹H → ²H + e⁺ + ν_e (×2)	Weak interaction: extremely slow, hence the Sun’s 10-billion-year life
2	²H + ¹H → ³He + γ (×2)	Fast; deuterium is consumed immediately
3	³He + ³He → ⁴He + 2 ¹H	Releases 12.86 MeV

4 ¹H → ⁴He + 2e⁺ + 2ν_e + 26.7 MeV

The positrons annihilate with electrons, and the 26.7 MeV figure includes that energy. Neutrinos escape the Sun in about two seconds and carry away roughly 0.5 MeV each. The energy in the form of photons takes tens of thousands of years to random-walk out of the dense core.

Hydrogen burning II: the CNO cycle

In stars more massive than about 1.3 solar masses (core T above ~1.7 × 10⁷ K), carbon, nitrogen and oxygen act as catalysts. The net result is the same, 4 protons → ⁴He + energy, but the reaction rate is far more sensitive to temperature, so it dominates in hotter stars.

Helium burning: the triple-alpha process

When hydrogen in the core is exhausted, the core contracts to about 10⁸ K and three helium nuclei fuse into carbon (via unstable ⁸Be). Further reactions create oxygen:

3 × ⁴He → ¹²C + γ | ¹²C + ⁴He → ¹⁶O + γ

5. Controlled Thermonuclear Fusion: Building a Star on Earth

The Sun uses gravity to confine its plasma and an enormous volume and time to make up for slow reactions. On Earth we have neither, so we must heat fuel to 10⁸ K and hold it long and dense enough that fusion outpaces energy loss. Three conditions must be met together:

High temperature (about 100–150 million K, ≈ 10–15 keV) so that nuclei can overcome the Coulomb barrier.
Sufficient density n so that collisions are frequent.
Long confinement time τ so that the plasma does not cool before enough energy is produced.

The Lawson criterion

Energy produced must exceed energy put in. For D–T this requires the product of density and confinement time to exceed a threshold:

n τ ≳ 10²⁰ m⁻³ s (and T ≈ 10⁸ K)

A more complete form is the triple product, n T τ ≳ 3 × 10²¹ keV s m⁻³. The fusion gain factor is Q = (fusion power out) / (heating power in). Q = 1 is breakeven; Q → ∞ is ignition, where alpha-particle heating alone sustains the burn.

Two roads to confinement

	Magnetic confinement	Inertial confinement
Idea	Charged plasma is trapped by strong magnetic fields (no material can touch 10⁸ K plasma)	A tiny fuel pellet is compressed and heated by lasers or beams in nanoseconds
Density / time	Low density (~10²⁰ m⁻³), long time (seconds to minutes)	Extremely high density, time ≈ 10⁻¹⁰ s
Devices	Tokamak (doughnut-shaped, toroidal + poloidal fields), stellarator	Laser facilities such as NIF (USA)
Flagship	ITER, Cadarache, France: designed for Q = 10 (≈ 500 MW from 50 MW input)	NIF achieved target gain above 1 in December 2022 (≈ 3.15 MJ out for ≈ 2.05 MJ laser energy)

Why isn’t fusion powering our grid yet?

Heating and stability: plasma develops turbulence and instabilities that leak heat.
Materials: 14 MeV neutrons damage reactor walls, so new structural materials are needed.
Tritium supply: tritium is scarce and radioactive (half-life ≈ 12.3 y), so reactors must breed it from lithium in a blanket: n + ⁶Li → ⁴He + ³H.
Engineering: net electricity requires Q well above 1 and efficient conversion; ITER’s schedule has been revised several times.

The reward is enormous: deuterium is abundant in seawater, there is no chain reaction (so no meltdown), no CO₂ emission, and the waste is far less long-lived than fission waste.

6. One-Page Formula & Fact Sheet

Concept	Formula / Value
Mass–energy	E = Δm c²; 1 u = 931.5 MeV/c²
Mass defect	Δm = Z m_p + (A−Z) m_n − M
B.E. per nucleon	B/A = Δm × 931.5 / A (MeV)
Q-value	Q = (Σ m_reactants − Σ m_products) × 931.5 MeV
Radius of nucleus	R = R₀ A^1/3, R₀ ≈ 1.2 fm
Coulomb barrier	U = e²/(4πε₀ r) = 1.44 MeV·fm / r(fm)
1 eV	1.602 × 10⁻¹⁹ J; 1 MeV = 1.602 × 10⁻¹³ J
Avogadro	N_A = 6.022 × 10²³ mol⁻¹
Fission energy	≈ 200 MeV per ²³⁵U; ≈ 8.2 × 10¹³ J per kg
Multiplication factor	k = 1 critical; N = N₀ kⁿ after n generations
D–T fusion	Q = 17.6 MeV; n gets 14.1 MeV, α gets 3.5 MeV
p–p chain	4 ¹H → ⁴He + 26.7 MeV
Lawson criterion	nτ ≳ 10²⁰ m⁻³ s, T ≈ 10⁸ K
Power from mass loss	P = (dm/dt) c²

7. Frequently Asked Questions (FAQs)

Q1. Why does fusion release more energy per kilogram than fission?

The B/A curve is steepest at the light end. Fusing deuterium and tritium raises B/A from about 1–3 MeV to 7 MeV (a jump of ~4–5 MeV per nucleon), while fission raises it from 7.6 to 8.5 MeV (~0.9 MeV per nucleon). Light fuels also have fewer nucleons per nucleus, so more reactions occur per kilogram.

Q2. Can a nuclear reactor explode like an atom bomb?

No. Reactor fuel is only 0.7–5% ²³⁵U, whereas a bomb needs > 90% enriched material assembled rapidly. A reactor can overheat (meltdown) or suffer a steam/chemical explosion if cooling fails, but it cannot undergo a nuclear explosion.

Q3. What is critical mass?

It is the minimum mass of fissile material for which neutron production balances loss, so that k ≥ 1. It depends on the material, its shape, purity and any neutron reflector around it.

Q4. If the Sun’s core is only 15 million K, how does fusion happen there?

Three reasons: (1) quantum tunnelling lets protons pass through the Coulomb barrier at far lower energy than classical physics demands; (2) a few particles in the high-energy tail of the Maxwell–Boltzmann distribution are fast enough; (3) the core is enormous and dense, with a billion-year timescale.

Q5. Why is iron the end of stellar fusion?

Iron-56 (with nickel-62) has the highest binding energy per nucleon. Fusing it, or splitting it, would require energy rather than release it, so the core stops generating pressure and collapses.

Q6. Why is a moderator needed in a reactor?

Fission neutrons are fast (~2 MeV) and are poorly captured by ²³⁵U. A moderator of light nuclei slows them to thermal energies (~0.025 eV) through elastic collisions, which raises the fission probability enormously.

Q7. Why are control rods made of cadmium or boron?

These elements have a very large neutron-absorption cross-section. Pushing the rods in absorbs neutrons and lowers k; pulling them out raises k.

Q8. What does the Q-factor in fusion mean?

Q is the ratio of fusion power produced to external heating power supplied to the plasma. Q = 1 is breakeven; ITER aims for Q = 10.

Q9. Is fusion waste-free and meltdown-proof?

Not entirely waste-free: neutrons activate reactor structures, and tritium must be handled carefully. But the waste is mostly short-lived compared with fission products, and a fault simply cools the plasma and stops the reaction. There is no chain reaction to run away.

Q10. What is the difference between fissile and fertile materials?

Fissile isotopes (²³⁵U, ²³⁹Pu, ²³³U) undergo fission with slow neutrons. Fertile isotopes (²³⁸U, ²³²Th) cannot sustain fission themselves but convert into fissile ones after neutron capture.

Q11. Why is heavy water a better moderator than ordinary water?

Deuterium absorbs far fewer neutrons than hydrogen, so more neutrons survive to cause fission. This allows reactors to operate on natural, unenriched uranium.

Q12. When will commercial fusion arrive?

No one can promise a date. Experimental devices have reached breakeven in inertial confinement (NIF) and long-duration plasmas in tokamaks, but a net-electricity power plant still needs advances in materials, tritium breeding and engineering. Most roadmaps point to the 2040s or later.

8. Multiple Choice Questions (MCQs)

Choose the single best answer. The answer key with explanations follows the questions.

Q1. The binding energy per nucleon is maximum for nuclei of mass number around:

(A) A ≈ 4

(B) A ≈ 56

(D) A ≈ 238

Q2. The approximate energy released in the fission of one ²³⁵U nucleus is:

(A) 2 MeV

(B) 20 MeV

(D) 2000 MeV

Q3. The role of a moderator in a nuclear reactor is to:

(A) Absorb neutrons

(B) Slow down fast neutrons

(D) Shield radiation

Q4. Which of the following is used as a control rod material?

(A) Graphite

(B) Heavy water

(D) Lead

Q5. A reactor operating at steady power has a multiplication factor k equal to:

(A) 0

(B) Less than 1

(D) Greater than 1

Q6. Fusion reactions require extremely high temperatures because:

(A) Nuclei are neutral

(B) Nuclei must overcome mutual Coulomb repulsion

(D) Gamma rays must be absorbed

Q7. The main source of the Sun’s energy is:

(A) Fission of uranium

(B) Chemical burning

(D) Gravitational contraction only

Q8. In D–T fusion (²H + ³H → ⁴He + n + 17.6 MeV), the neutron carries approximately:

(A) 20% of the energy

(B) 50% of the energy

(D) 100% of the energy

Q9. The Lawson criterion for D–T fusion is approximately:

(A) nτ ≈ 10¹⁰ m⁻³s

(B) nτ ≈ 10¹⁵ m⁻³s

(D) nτ ≈ 10³⁰ m⁻³s

Q10. Which of the following is a fertile material?

(A) ²³⁵U

(B) ²³⁹Pu

(D) ²³³U

Q11. Heavy water is preferred as a moderator because it:

(A) Is cheaper than graphite

(B) Absorbs very few neutrons while slowing them

(D) Is radioactive

Q12. A mass defect of 0.2 u corresponds to an energy of about:

(A) 18.6 MeV

(B) 93 MeV

(D) 931 MeV

Q13. In a tokamak, plasma is confined by:

(A) Gravitational field

(B) Combined toroidal and poloidal magnetic fields

(D) Material walls

Q14. The triple-alpha process in red giant cores produces:

(A) Helium-3

(B) Oxygen-16

(D) Iron-56

Q15. Which statement about fusion is INCORRECT?

(A) It releases more energy per kg than fission

(B) It needs temperatures of ~10⁸ K

(D) Deuterium is abundant in seawater

Q16. Assertion (A): Fission of heavy nuclei releases energy. Reason (R): Binding energy per nucleon of the fragments is greater than that of the parent nucleus.

(A) Both A and R true; R explains A

(B) Both true; R does not explain A

(D) A false, R true

MCQ Answer Key with Explanations

Q	Ans	Explanation
1	B	The B/A curve peaks near ⁵⁶Fe at ≈ 8.8 MeV.
2	C	≈ 0.9 MeV gain per nucleon × 235 ≈ 200 MeV.
3	B	Light nuclei slow fast neutrons to thermal energies.
4	C	Cadmium (and boron) strongly absorb neutrons.
5	C	k = 1 means critical: each fission leads to exactly one more.
6	B	Positively charged nuclei repel; high kinetic energy is needed to approach within ~10⁻¹⁵ m.
7	C	Hydrogen is fused to helium via the p–p chain.
8	C	Momentum conservation gives the lighter neutron (mass 1) 4/5 of the energy.
9	C	Typical value nτ ≳ 10²⁰ m⁻³s at T ≈ 10⁸ K.
10	C	²³²Th converts into fissile ²³³U after neutron capture.
11	B	Low neutron absorption allows natural uranium as fuel.
12	C	0.2 × 931.5 ≈ 186 MeV.
13	B	A toroidal field plus the field of the plasma current twist the field lines into a helix.
14	C	3 ⁴He → ¹²C at about 10⁸ K.
15	C	Fusion is thermally sustained and has no neutron-driven chain reaction.
16	A	Fragments are more tightly bound, so the extra binding energy is released.

9. Previous-Year-Question (PYQ) Style Practice

PYQ 1 [JEE Main pattern]

A nucleus of mass number 240 and binding energy per nucleon 7.6 MeV splits into two fragments of mass number 120 each, each with binding energy per nucleon 8.5 MeV. The energy released in the process is:

(A) 0.9 MeV (B) 9 MeV (C) 216 MeV (D) 240 MeV

PYQ 2 [NEET pattern]

Fusion reactions occur only at very high temperatures because:

(A) Nuclei are charged and repel each other (B) Nuclei are very heavy (C) Neutrons are unstable (D) Fuel is a gas

PYQ 3 [NEET pattern]

In the fission of ²³⁵U, approximately how much energy is released per nucleon of the original nucleus?

(A) 0.85 MeV (B) 8.5 MeV (C) 85 MeV (D) 200 MeV

PYQ 4 [JEE Main pattern]

In a nuclear reactor generating 1 MW of thermal power, how many fissions occur per second if each releases 200 MeV?

(A) 3.1 × 10¹⁶ (B) 3.1 × 10¹⁰ (C) 6.2 × 10¹² (D) 1.6 × 10¹⁹

PYQ 5 [JEE Main pattern]

Binding energy per nucleon of ²H and ⁴He are 1.1 MeV and 7.0 MeV. Energy released when two deuterons fuse to form one ⁴He nucleus is:

(A) 23.6 MeV (B) 28 MeV (C) 4.4 MeV (D) 11.8 MeV

PYQ 6 [JEE / NEET pattern]

If 0.1% of the mass of 1 kg of a substance is converted into energy, the energy released is (c = 3 × 10⁸ m/s):

(A) 9 × 10¹³ J (B) 9 × 10¹⁶ J (C) 9 × 10¹⁰ J (D) 9 × 10¹⁹ J

PYQ 7 [CBSE Board (3 marks)]

Write two differences between nuclear fission and fusion. Why is controlled fusion on Earth difficult?

PYQ 8 [CBSE / TN Board (2 marks)]

What is the function of control rods and moderator in a nuclear reactor?

Answers for the PYQ

1.Answer: (C). Q = total B.E. of fragments − B.E. of parent = 240 × 8.5 − 240 × 7.6 = 240 × 0.9 = 216 MeV.

2.Answer: (A). Kinetic energy must be high enough to overcome the electrostatic (Coulomb) repulsion so the nuclei come within the range of the strong nuclear force.

3.Answer: (A). 200 MeV / 235 ≈ 0.85 MeV per nucleon.

4.Answer: (A). Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J. Rate = 10⁶ / 3.2 × 10⁻¹¹ ≈ 3.1 × 10¹⁶ s⁻¹.

5.Answer: (A). B.E.(⁴He) = 4 × 7.0 = 28 MeV; B.E. of 2 deuterons = 2 × (2 × 1.1) = 4.4 MeV. Q = 28 − 4.4 = 23.6 MeV.

6.Answer: (A). Δm = 0.001 kg; E = Δm c² = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ J.

7.Model answer: (i) Fission splits a heavy nucleus (e.g. ²³⁵U) into lighter fragments, releasing ≈ 200 MeV, whereas fusion joins light nuclei (e.g. D + T) releasing 17.6 MeV. (ii) Fission is triggered by neutrons and runs as a chain reaction at ordinary temperatures; fusion requires ~10⁸ K and has no chain reaction. Controlled fusion is difficult because plasma must be heated to ~10⁸ K, held at sufficient density for a long enough time (Lawson criterion), and no material container can withstand such temperatures, so magnetic or inertial confinement is needed.

8.Model answer: The moderator (heavy water, graphite) slows fast fission neutrons to thermal energies so they can cause further fission. Control rods (cadmium, boron) absorb excess neutrons, keeping the multiplication factor k = 1 so the reaction proceeds at a steady rate; fully inserting them shuts down the reactor.

10. Numericals with Step-by-Step Solutions

Constants: 1 u = 931.5 MeV/c², 1 MeV = 1.602 × 10⁻¹³ J, N_A = 6.022 × 10²³, c = 3 × 10⁸ m/s. Atomic masses (u): ¹H 1.007825, n 1.008665, ²H 2.014102, ³H 3.016049, ³He 3.016029, ⁴He 4.002603.

Part A: Fully solved

Numerical 1: Binding energy of helium-4

Calculate the binding energy and binding energy per nucleon of ⁴He.

Mass of constituents = 2(1.007825) + 2(1.008665) = 4.031980 u

Δm = 4.031980 − 4.002603 = 0.029377 u

B.E. = 0.029377 × 931.5 = 27.36 MeV

B/A = 27.36 / 4 = 6.84 MeV per nucleon (hand-calculation using atomic masses)

Answer: B.E. ≈ 27.4 MeV; B/A ≈ 6.8 MeV (accepted value ≈ 7.07 MeV)

Numerical 2: Energy from 1 g of ²³⁵U

Find the energy released if 1 g of ²³⁵U undergoes complete fission at 200 MeV per fission.

Number of atoms = (1/235) × 6.022 × 10²³ = 2.563 × 10²¹

Energy per fission = 200 × 1.602 × 10⁻¹³ = 3.204 × 10⁻¹¹ J

Total E = 2.563 × 10²¹ × 3.204 × 10⁻¹¹ = 8.2 × 10¹⁰ J

In practical units: 8.2 × 10¹⁰ J / 3.6 × 10⁶ ≈ 22,800 kWh ≈ 0.95 MW·day

Answer: ≈ 8.2 × 10¹⁰ J (about 1 MW for one day)

Numerical 3: Fuel use of a reactor

A reactor produces 200 MW of thermal power by fission of ²³⁵U (200 MeV per fission). Find the fission rate and the mass of ²³⁵U consumed per day.

Energy per fission = 3.2 × 10⁻¹¹ J

Rate = 200 × 10⁶ / 3.2 × 10⁻¹¹ = 6.25 × 10¹⁸ fissions/s

Per day: 6.25 × 10¹⁸ × 86400 = 5.4 × 10²³ fissions

Moles = 5.4 × 10²³ / 6.022 × 10²³ = 0.897 mol; mass = 0.897 × 235 = 211 g

Answer: 6.25 × 10¹⁸ fissions/s; ≈ 0.21 kg of ²³⁵U per day

Numerical 4: Q-value of D–T fusion

Calculate the energy released in ²H + ³H → ⁴He + n, and the share carried by the neutron.

Reactants: 2.014102 + 3.016049 = 5.030151 u

Products: 4.002603 + 1.008665 = 5.011268 u

Δm = 0.018883 u; Q = 0.018883 × 931.5 = 17.59 MeV

Momentum conservation: p_n = p_α, so K ∝ 1/m. Neutron share = 4/(4+1) = 80%

Answer: Q ≈ 17.6 MeV; neutron ≈ 14.1 MeV, alpha ≈ 3.5 MeV

Numerical 5: Coulomb barrier and temperature for D–D fusion

Estimate the Coulomb barrier for two deuterons just touching (R = 1.2 A^1/3 fm) and the classical temperature needed.

R = 1.2 × 2^1/3 = 1.51 fm; separation r = 2R = 3.02 fm

U = 1.44 MeV·fm / 3.02 fm = 0.477 MeV

Set 2 × (3/2) k T = U, so T = U / 3k = (0.477 × 10⁶ × 11604 K) / 3 ≈ 1.8 × 10⁹ K

Answer: Barrier ≈ 0.48 MeV; T ≈ 2 × 10⁹ K classically. Real fusion occurs at ~10⁸ K thanks to tunnelling and the high-energy tail

Numerical 6: The Sun is losing mass

The Sun radiates 3.8 × 10²⁶ W. Find the mass lost per second.

Use P = (dm/dt) c²

dm/dt = 3.8 × 10²⁶ / (3 × 10⁸)² = 3.8 × 10²⁶ / 9 × 10¹⁶

= 4.2 × 10⁹ kg/s

Answer: ≈ 4.2 × 10⁹ kg/s (about 4 million tonnes every second)

Numerical 7: Hydrogen consumption in the Sun

Using 26.7 MeV per ⁴He formed, find (a) the number of p–p cycles per second and (b) the hydrogen consumed per second in the Sun (L = 3.8 × 10²⁶ W).

Energy per cycle = 26.7 × 1.602 × 10⁻¹³ = 4.28 × 10⁻¹² J

(a) Rate = 3.8 × 10²⁶ / 4.28 × 10⁻¹² = 8.9 × 10³⁷ cycles/s

(b) 4 protons per cycle: 4 × 1.673 × 10⁻²⁷ kg × 8.9 × 10³⁷ = 5.9 × 10¹¹ kg/s

Answer: ≈ 9 × 10³⁷ cycles/s; ≈ 6 × 10¹¹ kg/s of hydrogen (≈ 600 million tonnes/s)

Numerical 8: Growth of a chain reaction

A reactor starts with 1000 neutrons and k = 1.2. How many neutrons are there after 5 generations? What if k = 0.9?

N = N₀ kⁿ

k = 1.2: N = 1000 × (1.2)⁵ = 1000 × 2.488 = 2488

k = 0.9: N = 1000 × (0.9)⁵ = 1000 × 0.590 = 590

Answer: ≈ 2490 neutrons (supercritical growth); ≈ 590 neutrons (subcritical decay)

Numerical 9: Fission vs fusion per kilogram

Compare the energy per kilogram from ²³⁵U fission (200 MeV per atom) and from D–T fusion (17.6 MeV per ²H + ³H pair).

Fission: 8.2 × 10¹³ J/kg (from Numerical 2 scaled: 8.2 × 10¹⁰ J/g)

Fusion: one pair has mass ≈ 5 u = 5 × 1.66 × 10⁻²⁷ = 8.3 × 10⁻²⁷ kg

Energy = 17.6 × 1.602 × 10⁻¹³ = 2.82 × 10⁻¹² J per pair

Per kg: 2.82 × 10⁻¹² / 8.3 × 10⁻²⁷ = 3.4 × 10¹⁴ J/kg

Ratio = 3.4 × 10¹⁴ / 8.2 × 10¹³ ≈ 4.1

Answer: Fusion yields ≈ 4 times more energy per kg than fission

Numerical 10: Lawson criterion in use

A fusion plasma has density n = 5 × 10¹⁹ m⁻³. Find the minimum energy confinement time required if nτ ≥ 10²⁰ m⁻³s.

τ ≥ 10²⁰ / (5 × 10¹⁹)

τ ≥ 2 s

Answer: Confinement time ≳ 2 seconds

Quick Revision & Final Thoughts

Five lines to remember

Energy comes from mass defect: E = Δm c², and the B/A curve peaks at iron.
Fission (≈ 200 MeV) is controlled by moderators and control rods at k = 1.
Fusion (D–T 17.6 MeV) needs ~10⁸ K, with the Coulomb barrier as the villain and quantum tunnelling as the helper.
Stars run on the p–p chain, CNO cycle and triple-alpha process until the core turns to iron.

Controlled fusion needs n τ ≳ 10²⁰ m⁻³s at 10⁸ K: tokamak, stellarator or laser implosion, with ITER and NIF leading the way.

Atoms and Nuclei – Master Hub

Nuclear Energy – From Splitting Atoms to Bottling a Star