A full lesson: a cartoon movie to start, then how joining light nuclei releases energy, why fusion needs a temperature of millions of kelvin, how to check A, Z and the Q value of any nuclear reaction, the proton–proton cycle that powers the sun, the sun's future and fusion on Earth, NEET shortcuts, problems to solve yourself, and 12 MCQs.
The sun is not a coal fire. In its core, at about 1.5 × 10⁷ K, hydrogen nuclei fuse into helium. Four protons make one helium-4 nucleus and release 26.7 MeV, and that energy reaches us as sunlight.
On a Sunday morning Priya puts a pot of rice in her solar cooker on the terrace. By lunch it is cooked, with no gas and no firewood. Arjun looks at the sun and asks: "Is the sun a giant ball of burning coal? When will it run out?" Good question. A sun made of coal, burning as brightly as ours, would be finished in a few thousand years. Yet the sun has been shining for about 5 billion years.
The secret is nuclear fusion. In the core of the sun, light nuclei (mostly protons) join to make heavier ones, climbing the binding energy curve from the light side. In the last lesson heavy nuclei released energy by splitting; today light nuclei release energy by joining.
"Every sunbeam that cooks Priya's rice started as four protons becoming one helium nucleus. The sunlight you see left the sun about eight minutes ago."
"Two protons both carry positive charge. What must happen before they can come close enough to fuse?"
They must climb over the Coulomb barrier: their repulsion. Only very fast protons can do that, so the gas must be extremely hot. That is why fusion by heating is called thermonuclear fusion.
When two light nuclei join, the product sits higher on the E_bn curve, so it is more tightly bound and energy is released. Pick a reaction to watch the nuclei merge, then slide the amount of fuel to see how much energy a little deuterium holds.
Two protons make a deuteron, a positron and a neutrino. This is the first step in the sun.
and ²H + ²H → ³H + ¹H + 4.03 MeV. Deuterium is found in ordinary sea water.
Fission of ²³⁵U gives about 0.85 MeV per nucleon. Fusing hydrogen into helium gives about eight times more per kilogram of fuel.
"Fission climbs the curve from the right, fusion climbs it from the left. And the left side is much steeper, so fusion gives far more energy per kilogram."
A proton approaching another proton must climb a hill of Coulomb potential energy before the short-range nuclear force can grab it. The top of the hill is about 400 keV. Heat the gas and watch an average proton, or launch one that is many times faster than average.
k = 1.38 × 10⁻²³ J K⁻¹ is Boltzmann's constant.
This is the NCERT estimate of the barrier for two protons.
Fusion by heating is thermonuclear fusion. The sun's core is only 1.5 × 10⁷ K, so in the sun fusion happens through the few protons whose energy is far above the average.
"The average proton in the sun never makes it. But there are so many protons that the rare very fast ones are enough to keep the sun shining. (A quantum effect called tunnelling also helps; that is beyond NCERT.)"
In every nuclear reaction the total nucleon number A and the total charge are conserved, so the top numbers (A) and the bottom numbers (Z, counting +1 for e⁺ and −1 for e⁻) must add up the same on both sides. (A proton can turn into a neutron, as in the first p–p step, so the proton and neutron counts are not separately fixed.) Choose a reaction, then find the unknown particle X with the two sliders. The scales level out when you are right, and then the Q value appears.
Electrons, positrons and neutrinos have A = 0. A positron has Z = +1, an electron Z = −1.
With masses in u, multiply by 931.5 MeV. Q > 0: energy released (exothermic). Q < 0: energy must be supplied.
When no proton turns into a neutron (or back), as in fission, d + d or d + t, the rest masses of the protons and neutrons are the same before and after. Then only the binding energy changes, and that difference is Q. (In the p–p step Q also includes the n–p mass difference and the positron mass.)
"A chemical equation balances atoms. A nuclear equation balances nucleons: top numbers and bottom numbers. Atoms of an element can change into atoms of another element, and that is allowed."
"Do chemical reactions also change mass, or is that only nuclear?"
Chemical reactions also change mass by E/c², but a chemical reaction gives only a few eV, so the mass change is about a million times smaller than in a nuclear reaction. It is too small to measure on a balance.
The sun burns hydrogen into helium in four steps. The first three steps happen twice (top and bottom lanes) to supply the two ³He nuclei needed in the fourth. Play the movie or step through it, and watch the particle counter and the energy add up.
| Step | Reaction (NCERT) | Energy |
|---|---|---|
| 1 (×2) | ¹H + ¹H → ²H + e⁺ + ν | 0.42 MeV |
| 2 (×2) | e⁺ + e⁻ → γ + γ | 1.02 MeV |
| 3 (×2) | ²H + ¹H → ³He + γ | 5.49 MeV |
| 4 | ³He + ³He → ⁴He + ¹H + ¹H | 12.86 MeV |
| Net | 4 ¹H + 2e⁻ → ⁴He + 2ν + 6γ | 26.7 MeV |
"Remember 2–2–6: two positrons (which meet two electrons), two neutrinos and six gamma photons for every helium nucleus. And the energy: 2 × (0.42 + 1.02 + 5.49) + 12.86 = 26.7 MeV."
Slide through the life of the sun, then switch to a fusion reactor on Earth, where the fuel is a plasma so hot that no wall can touch it. Magnets have to hold it.
It has enough hydrogen in the core for about another 5 billion years.
The core cools, shrinks under gravity and heats up; the outer layers swell into a red giant. Helium then fuses into carbon. Fusion stops at nuclei near the peak of the E_bn curve.
The fuel is a plasma of ions and electrons. Keeping it together (confinement) is the hardest problem. India is a partner in the international ITER project.
"In the sun, gravity does the confining for free. On Earth we must do it with magnetic fields, and that is why fusion power is still being built."
Fission is the opposite: heavy → two lighter. Both move towards the peak of E_bn (A ≈ 56) and both release energy.
K = 400 keV gives T ≈ 3 × 10⁹ K. Sun's core: 1.5 × 10⁷ K. Controlled fusion: about 10⁸ K.
Four protons are used; two of them become neutrons (that is where the positrons come from).
= [4 m(¹H) − m(⁴He)] × 931.5 = 0.0287 u × 931.5. About 0.7% of the mass becomes energy.
²H + ²H → ⁴He: 4 × 7.07 − 4 × 1.11 ≈ 23.8 MeV.
ΔA = 0, ΔZ = +1 ⇒ e⁺; ΔA = 1, ΔZ = 1 ⇒ p; ΔA = 4, ΔZ = 2 ⇒ α (⁴He); ΔA = 0, ΔZ = 0 ⇒ γ or ν.
"For fusion questions: find X by balancing, find Q by mass loss times 931.5, and if they ask temperature, use three halves k T."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Atomic masses: ¹H 1.007825 u, ²H 2.014102 u, ³H 3.016049 u, ³He 3.016029 u, ⁴He 4.002603 u, ⁶Li 6.015122 u, n 1.008665 u.
The Coulomb barrier for two protons is about 400 keV. Estimate the temperature at which the average proton in a gas has this kinetic energy. (k = 1.38 × 10⁻²³ J K⁻¹)
(3/2) k T = K, with K in joules.
Find the energy released in ²H + ²H → ³He + n.
Q = [2 m(²H) − m(³He) − m(n)] × 931.5 MeV.
Using atomic masses, find the energy released when four hydrogen atoms make one helium-4 atom. What fraction of the mass turns into energy?
Atomic masses already include the electrons, so 4 m(¹H) − m(⁴He) gives the net Q directly.
The sun radiates 3.8 × 10²⁶ W. Each helium nucleus made gives 26.7 MeV. How many helium nuclei form per second, and how much hydrogen is used per second? (m_H ≈ 1.67 × 10⁻²⁷ kg)
Number per second = P ÷ (26.7 × 1.6 × 10⁻¹³ J). Each needs 4 hydrogen nuclei.
In ⁶₃Li + ¹₀n → ³₁H + X, identify X and find Q.
Balance A and Z first. Then Q = [m(⁶Li) + m(n) − m(³H) − m(X)] × 931.5 MeV.
Take 1 u = 931.5 MeV/c², k = 1.38 × 10⁻²³ J K⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "source of the sun's energy" | fusion of hydrogen into helium | Choosing fission or "chemical burning" |
| "why fusion needs high temperature" | to overcome the Coulomb repulsion | Saying "to break the nuclei" or "to ionise atoms" |
| "temperature for barrier K" | (3/2) k T = K | Using k T = K, or leaving K in eV |
| "find X" in a reaction | Σ A and Σ Z balance | Forgetting that a positron has Z = +1 and A = 0 |
| "Q value" with masses | Δm (u) × 931.5 MeV | Subtracting in the wrong order and getting −Q |
| "net energy of the p–p cycle" | 26.7 MeV per ⁴He | Giving 12.86 MeV (only the last step) |
| "particles per helium in the sun" | 2 e⁺, 2 ν, 6 γ | Counting the steps once instead of twice |
| "what is conserved" | total nucleon number A and total charge (ΣA, ΣZ) | "number of atoms of each element" |
| "energy per unit mass, fusion vs fission" | fusion (H → He) ≈ 8 × more | Comparing energy per reaction (200 MeV > 26.7 MeV) |
This completes the Nuclei chapter. Try the Nuclei Challenge: it mixes nuclear size and density, mass defect, binding energy, fission and fusion, exactly as NEET does.
Values used: 1 u = 931.5 MeV/c², k = 1.38 × 10⁻²³ J K⁻¹, 1 MeV = 1.6 × 10⁻¹³ J, Coulomb barrier for two protons ≈ 400 keV (NCERT), sun's luminosity 3.8 × 10²⁶ W. Atomic masses from standard tables.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.