Nuclear Force and Fission · Vidya720
Physics · Class 12 Nuclei · NEET

The nuclear force and fission, from a village power cut to a NEET question

A full lesson: a cartoon movie to start, then the strongest glue in nature and its potential energy curve, why a short-range force keeps the binding energy per nucleon flat, how the binding energy curve tells us which reactions give energy, the fission of uranium-235 with its Q value, the chain reaction, NEET shortcuts, problems to solve yourself, and 12 MCQs.

NCERT Class 12 Physics Part II · Ch 13 Nuclei · §13.5 Nuclear force, §13.6 Nuclear energy, §13.6.1 Fission · pp. 313–315 · about 60 min

Voice
About 6 minutes. The animations start by themselves as the teacher explains them. Each stage's Listen button uses the same language.

One kilogram of coal gives about 10⁷ J; one kilogram of uranium gives about 10¹⁴ J. The secret is inside the nucleus: a neutron splits uranium-235 into two middle-sized nuclei, about 200 MeV comes out, and the new neutrons can split more nuclei.

Stage 1 · The hook (5 min)

The pellet that lights a city

Level

It is evening in Dadi's village and the power goes off. Kabir lights a candle and asks the obvious question: where does our electricity actually come from? Part of Karnataka's power comes from the Kaiga atomic power station, where there is no smoke and no coal train. The fuel is uranium, packed as pellets no bigger than a fingertip.

Here is the surprise. Burning coal is chemistry: electrons rearrange between atoms, and each reaction gives a few eV. Splitting a uranium nucleus is nuclear: the nucleons rearrange, and each event gives about 200 MeV, roughly a hundred million times more. In this lesson you will meet the force that holds a nucleus together and see exactly why splitting a heavy nucleus releases energy.

Teacher says

"Chemical energy lives in the electron cloud; nuclear energy lives in the nucleus. Same rule, E = mc², but MeV instead of eV: a million times per particle, and about 10⁸ times per fission event."

Pause and predict

"A nucleus packs many protons into a tiny space. They all repel each other. So why does the nucleus not fly apart?"

Answer (this is Stage 2)

A third force acts inside the nucleus: the nuclear force. It attracts protons and neutrons alike and, at nuclear distances, it is much stronger than the Coulomb repulsion between protons.

Stage 2 · The concept (10 min)

The nuclear force: the strongest glue

Level

Gravity between two protons is far too weak, and the Coulomb force pushes them apart. The nucleus is held by the nuclear force. Drag the right-hand nucleon (or use the slider) and watch the potential energy curve. Switch the pair to test whether the force cares about charge.

Much stronger

nuclear ≫ Coulomb ≫ gravity

At nuclear distances it beats the Coulomb repulsion between protons. Gravity is about 10³⁶ times weaker than the Coulomb force.

Short range

range: a few fm

Attractive from about 0.8 fm out to a few femtometres, then it dies away almost completely. 1 fm = 10⁻¹⁵ m.

Repulsive core

r < r₀ ≈ 0.8 fm: repulsive

The potential energy is least at r₀ ≈ 0.8 fm. Push closer and the force becomes strongly repulsive, so nucleons keep their distance.

Charge independent

n–n ≈ n–p ≈ p–p

The nuclear force is nearly the same for all three pairs. Only the p–p pair also feels the Coulomb push.

Teacher says

"Remember the curve, not just the words. Lowest point at about 0.8 fm. To the right of it, attraction that dies within a few fermi. To the left, a steep wall of repulsion."

Stage 3 · The law (10 min)

Short range means each nucleon has only a few partners

Level

A nucleon feels only its nearest neighbours, because the force dies within a few fm. Tap any nucleon to see who it can pull. Then grow the nucleus and compare with an imaginary long-range force, and with the Coulomb force between protons, which really is long range.

Saturation

pairs ∝ A ⇒ E_bn ≈ constant

Each nucleon bonds with a fixed few neighbours, so the total binding energy grows in proportion to A. That is why E_bn stays near 8 MeV for 30 < A < 170.

Long range would be different

pairs = A(A − 1)/2

If every nucleon pulled every other, E_bn would keep growing with A. It does not, so the force must be short range.

Coulomb spoils heavy nuclei

p–p pairs = Z(Z − 1)/2

Every proton repels every other proton. In heavy nuclei this grows like Z², so E_bn falls for A > 170. That fall is what makes fission possible.

Teacher says

"Short-range attraction saturates; long-range repulsion keeps adding up. Heavy nuclei lose that tug of war a little, and that small loss is the energy we get from fission."

Pause and predict

"In a nucleus of A = 200, roughly how many other nucleons does one nucleon in the middle really pull on: about 200, about 100, or about 12?"

Answer

About a dozen, its nearest neighbours. (The picture here is flat, so it shows 6.) That number does not change as the nucleus grows, which is exactly what saturation means.

Stage 4 · Maths (10 min)

Nuclear energy: read it off the binding energy curve

Level

The curve of binding energy per nucleon E_bn against A peaks near iron (A ≈ 56). Any process that moves nucleons up the curve gives out energy. Choose a process, drag the slider (or drag the red point) and read the energy released.

  1. Total binding energy

    E_b = A × E_bn

    The energy needed to pull the nucleus completely apart into free nucleons.

  2. Energy released

    Q = E_b(products) − E_b(reactants)

    The nucleon count stays the same, so the gain is simply the extra binding of the products.

  3. NCERT estimate for fission

    Q ≈ 240 × (8.5 − 7.6) ≈ 216 MeV

    A ≈ 240 at 7.6 MeV per nucleon splits into two A ≈ 120 at 8.5 MeV per nucleon: about 200 MeV per fission.

Teacher says

"Climb the hill and you get paid. Heavy nuclei climb by splitting, light nuclei climb by joining. Iron sits on top, so splitting iron or joining two medium nuclei costs energy instead."

Stage 5 · The big idea (10 min)

Fission of uranium-235, frame by frame

Level

A slow neutron is absorbed by ²³⁵U. The excited ²³⁶U wobbles like a liquid drop, forms a neck and splits into two middle-sized nuclei plus 2 to 4 neutrons. Play the movie or step through it. Change the fragment pair: there are many possible ways to split.

The NCERT equation

¹₀n + ²³⁵₉₂U → ²³⁶₉₂U → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3 ¹₀n

Top numbers: 1 + 235 = 144 + 89 + 3. Bottom numbers: 0 + 92 = 56 + 36 + 0.

Q value from masses

Q = [Σm(before) − Σm(after)] c²

With 1 u = 931.5 MeV/c². For the Ba–Kr split the mass loss is about 0.186 u, so Q ≈ 173 MeV at the moment of splitting.

About 200 MeV in all

fragments' KE + n + γ + β decays

Most of it is kinetic energy of the fragments, which becomes heat. The fragments are radioactive and give out more energy later (β and γ), bringing the total to about 200 MeV.

Enrichment (not in the current NCERT text): radioactivity in one minute

The fission fragments are neutron-rich, so they are radioactive: they change by themselves into other nuclei. NEET may still use the three words. α decay: a helium nucleus ⁴₂He is emitted, so A falls by 4 and Z by 2. β⁻ decay: a neutron changes into a proton, and an electron and an antineutrino are emitted, so A stays the same and Z rises by 1. γ decay: an excited nucleus emits a high-energy photon, so A and Z do not change. Half-life and decay laws are not part of the rationalised syllabus.

Teacher says

"Check every fission equation twice: top numbers add up, bottom numbers add up. The neutron has top 1 and bottom 0, and that is where most students slip."

Stage 6 · Real life (5 min)

The chain reaction: a bomb or a power station

Level

Each fission gives about 2.5 neutrons on average. If more than one of them causes another fission, the number of fissions grows generation after generation. Fire a neutron into this lump of uranium, then push the control rods in or pull them out.

Multiplication factor

k = fissions in one generation ÷ fissions in the one before

k > 1: the reaction grows (uncontrolled, as in a bomb). k = 1: steady (a reactor). k < 1: it dies out.

Control rods

cadmium or boron soak up neutrons

Pushing them in lowers k. Operators keep k = 1 so the power stays steady. (Reactor details are beyond the current NCERT text.)

Small lumps fizzle

too small ⇒ neutrons escape

In a small lump most neutrons leave through the surface before they hit a nucleus, so k stays below 1. Slow neutrons split ²³⁵U best, so reactors use a moderator such as water or graphite to slow them.

Teacher says

"The same physics runs a power station and a bomb. The only difference is control: in a reactor every fission causes exactly one more, on average."

Stage 7 · NEET level (10 min)

Shortcuts and tricks

Level

Order of the forces

F_nuclear > F_Coulomb ≫ F_gravity

For two protons 1 fm apart: Coulomb ≈ 230 N, gravity ≈ 1.9 × 10⁻³⁴ N. Ratio ≈ 10³⁶.

Three numbers for the curve

r₀ ≈ 0.8 fm, range ≈ few fm

r < r₀ repulsive, r > r₀ attractive, r beyond a few fm: negligible.

Q from binding energy

Q = ΣA·E_bn (after) − ΣA·E_bn (before)

Positive Q means energy is released. Free neutrons have zero binding energy.

Q from masses

Q = Δm × 931.5 MeV (Δm in u)

Use atomic masses on both sides; the electrons cancel when Z balances.

Per fission, per nucleon

≈ 200 MeV, ≈ 0.85 MeV per nucleon

200 MeV = 3.2 × 10⁻¹¹ J. One kilogram of ²³⁵U ≈ 8 × 10¹³ J, against about 10⁷ J for coal.

Fissions per second

N = P ÷ (3.2 × 10⁻¹¹ J)

A 1 W heater needs 3.1 × 10¹⁰ fissions per second. A 1000 MW plant: 3.1 × 10¹⁹ per second, about 1 kg of ²³⁵U per day.

Teacher says

"Four tools solve almost every fission question: balance A and Z, Q from E_bn, Q from mass loss times 931.5, and power divided by 200 MeV."

Stage 8 · Solve it yourself

Five worked problems

Level

Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute.

Problem 1 · Comparing forces

Two protons are 1 fm apart. Find the Coulomb force and the gravitational force between them, and their ratio. (k = 9 × 10⁹ N m² C⁻², G = 6.67 × 10⁻¹¹ N m² kg⁻², m_p = 1.67 × 10⁻²⁷ kg.)

Hint

F = k e²/r² and F = G m²/r², with r = 10⁻¹⁵ m.

Full solution
F_C = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² / (10⁻¹⁵)² ≈ 230 N
F_G = 6.67 × 10⁻¹¹ × (1.67 × 10⁻²⁷)² / (10⁻¹⁵)² ≈ 1.9 × 10⁻³⁴ N
F_C / F_G ≈ 1.2 × 10³⁶
230 N against 1.9 × 10⁻³⁴ N. The nuclear force must beat the 230 N to hold the protons.
Problem 2 · Balancing a fission

Find x and y in ¹₀n + ²³⁵₉₂U → ¹⁴⁰₅₄Xe + ⁹⁴ₓSr + y ¹₀n.

Hint

Add the top numbers, then add the bottom numbers. A neutron has A = 1, Z = 0.

Full solution
Z: 0 + 92 = 54 + x + 0 ⇒ x = 38
A: 1 + 235 = 140 + 94 + y ⇒ y = 2
x = 38 (strontium), y = 2 neutrons
Problem 3 · Energy from the curve

A nucleus with A = 240 and E_bn = 7.6 MeV splits into two nuclei of A = 120 with E_bn = 8.5 MeV. Find the energy released.

Hint

Q = (binding energy after) − (binding energy before).

Full solution
before: 240 × 7.6 = 1824 MeV
after: 2 × 120 × 8.5 = 2040 MeV
Q = 2040 − 1824 = 216 MeV
≈ 216 MeV, about 200 MeV per fission
Problem 4 · Q value from masses

Find Q for ¹₀n + ²³⁵₉₂U → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n. Masses: ²³⁵U = 235.043930 u, ¹⁴¹Ba = 140.914411 u, ⁹²Kr = 91.926156 u, n = 1.008665 u. Take 1 u = 931.5 MeV/c².

Hint

One neutron cancels on both sides, so effectively ²³⁵U → Ba + Kr + 2n.

Full solution
before: 235.043930 + 1.008665 = 236.052595 u
after: 140.914411 + 91.926156 + 3 × 1.008665 = 235.866562 u
Δm = 0.186033 u
Q = 0.186033 × 931.5 ≈ 173.3 MeV
Q ≈ 173 MeV (the fragments' later decays add about 25 MeV more)
Problem 5 · Fuel for a power station

A reactor produces 1000 MW of heat. Each fission of ²³⁵U gives 200 MeV. How many fissions happen per second, and what mass of ²³⁵U is used in one day?

Hint

200 MeV = 200 × 1.6 × 10⁻¹³ J. Then moles = number ÷ N_A, and 1 mol of ²³⁵U is 235 g.

Full solution
E per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J
fissions per second = 10⁹ / 3.2 × 10⁻¹¹ ≈ 3.1 × 10¹⁹
per day: 3.1 × 10¹⁹ × 86 400 ≈ 2.7 × 10²⁴ nuclei
mass = 2.7 × 10²⁴ / 6.02 × 10²³ × 235 g ≈ 1.05 kg
3.1 × 10¹⁹ fissions per second, about 1 kg of ²³⁵U a day
Stage 9 · Practice ladder

12 MCQs, from easy to NEET-level

Level

Take 1 u = 931.5 MeV/c², 1 MeV = 1.6 × 10⁻¹³ J, N_A = 6.02 × 10²³ mol⁻¹. Write the formula first, then click.

Score: 0 / 12

Stage 10 · Exam corner

How NEET asks this

If the question says…Write this firstThe trap
"nature of the nuclear force"strong, short range, charge independentCalling it long range or "only between protons"
"two nucleons closer than 0.8 fm"repulsiveSaying the nuclear force is attractive at all distances
"force between n–n, n–p, p–p"nuclear part nearly equalForgetting that p–p also has Coulomb repulsion
"why E_bn is nearly constant"saturation: short rangeAnswering "because the nuclear force is strong"
"energy released when A splits / joins"Q = E_b(after) − E_b(before)Multiplying E_bn by only one fragment's A
"find x neutrons" in a fission equationΣ A and Σ Z both balanceCounting the incoming neutron on one side only
"Q value" with atomic massesΔm (u) × 931.5 MeVUsing 931.5 with Δm in kg, or forgetting the extra neutrons
"fissions per second for P watts"P ÷ (200 × 1.6 × 10⁻¹³)Leaving 200 MeV in MeV and getting a 10¹³ error
"steady chain reaction"k = 1Choosing k > 1 because "it must keep going"

Mistakes to avoid

Where this connects next

Lesson 5 climbs the curve from the other side: light nuclei joining in fusion, the energy of the sun, why fusion needs millions of kelvin, and how A, Z and Q are checked in any nuclear reaction.

Values used: 1 u = 931.5 MeV/c² = 1.66 × 10⁻²⁷ kg, 1 MeV = 1.6 × 10⁻¹³ J, e = 1.6 × 10⁻¹⁹ C, N_A = 6.02 × 10²³ mol⁻¹. The E_bn curve is a smooth liquid-drop fit for A ≥ 17 and measured values for light nuclei. The nuclear potential curve is schematic in shape.

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