Take one slow breath before you start. Nuclei has a reputation for being the strange, scary corner of Class 12 Physics — all that talk of mass turning into energy, of bombs and stars — and yet it is quietly one of the most forgiving chapters in the whole book. The official CBSE 2026-27 syllabus for this chapter asks for six things, the arithmetic almost never goes beyond a subtraction and a multiplication, and every single conceptual question in it can be answered by looking at one graph. Not a set of graphs. One.
That graph is the binding energy per nucleon curve, and it is going to be the spine of this entire page. We will draw it early, from real computed numbers, and then every later question — why does nuclear fission release energy, why does nuclear fusion release energy, why is iron the end of the road for a star, why is helium-4 oddly stable, why are heavy nuclei unstable in the first place — will be answered by pointing back at the same picture. If you learn to read that one curve properly, you have learnt the chapter. Everything else is bookkeeping.
There is one more thing I want to settle straight away, because it is the single biggest source of panic in this chapter. If you open an older textbook or a random set of notes, you will find pages on radioactivity, on α, β and γ decay, on the radioactive decay law, on half-life and mean life, and on the nuclear reactor. None of that is in the 2026-27 CBSE syllabus for Nuclei. I have not deleted it — there is an honest, clearly-labelled amber section near the end that tells you exactly what those topics are and where they still matter (JEE, NEET, olympiads, general curiosity). But do not spend your exam preparation there, and do not feel behind because someone else is memorising decay constants.
Read this page slowly, with a pen. Every worked example is designed to be redone on blank paper afterwards, and they are deliberately arranged as a staircase: one-step mass defect first, then binding energy per nucleon, then full exam-style fission and fusion energy balances. If you can climb that staircase without help, the exam questions on this chapter will look like old friends.
Meet Your Tutor
Nuclear Physics becomes manageable when you keep mass, energy and stability in one picture. I will help you organise mass defect, binding energy, decay and fission-fusion questions around the binding-energy curve, with a unit check before every numerical answer.
What You’ll Learn
- Composition of the Nucleus: Protons, Neutrons and the A, Z Notation
- The Atomic Mass Unit u and Why 1 u = 931.5 MeV Matters
- Size of the Nucleus: R = R₀A1/3 and Why Nuclear Density Ignores A
- How Small Is Small? The Nucleus-and-Atom Scale Gap
- Nuclear Force: Every Feature CBSE Asks You to List
- Mass-Energy Relation E = mc² and Mass Defect
- Binding Energy and Binding Energy Per Nucleon
- Binding Energy Per Nucleon Curve Explained: Every Conclusion in One Graph
- The Iron Hill Rule: One Memory Device for the Whole Chapter
- Nuclear Fission: Where the 200 MeV Actually Comes From
- Nuclear Fusion, the Coulomb Barrier and Why Stars Shine
- Why Fission AND Fusion Both Release Energy — The One Unifying Idea
- Nuclear Fission and Fusion Difference Class 12: Side-by-Side Table
- Not in Your 2026-27 Exam: Radioactivity, Decay Law, Half-Life, Reactor
- Nuclei Class 12 Physics Important Questions: What Examiners Actually Ask
- Common Mistakes and Unit Traps in Nuclei Numericals
- Practice Worksheet (12 questions with fully worked answers)
Your Game Plan
Six short sittings. Do them in this order — the chapter is built like a staircase and skipping a step is what makes it feel hard.
- Get the vocabulary straight (20 minutes). Nucleon, proton number Z, neutron number N, mass number A, and the notation. Nothing to calculate yet.
- Make 1 u = 931.5 MeV automatic (20 minutes). This single conversion carries about half the marks in the chapter.
- Size and density (30 minutes). R = R₀A1/3, and the surprise that falls out of it — every nucleus has the same density.
- Mass defect → binding energy → binding energy per nucleon (60 minutes). Three ideas, one arithmetic chain. Do every worked example on paper.
- Live inside the curve (45 minutes). Draw the binding energy per nucleon curve from memory until the shape is in your hand, not just your head.
- Fission and fusion as consequences (45 minutes). By this point you should be able to predict both answers before you read them.
Study Notes
Composition of the Nucleus: Protons, Neutrons and the A, Z Notation
Start with the picture you already have. An atom is a tiny, heavy core with electrons around it. That core is the nucleus, and it is made of exactly two kinds of particle: protons, which carry a positive charge of +e each, and neutrons, which carry no charge at all. Because both of them live inside the nucleus and behave almost identically as far as the nuclear force is concerned, we give them a single family name: nucleons.
Three counting numbers describe any nucleus completely, and CBSE expects you to use them fluently:
- Atomic number Z — the number of protons. This is also the number of electrons in the neutral atom, so Z is what decides which element you are looking at. Change Z and you have changed the chemistry.
- Neutron number N — the number of neutrons. Change N alone and you still have the same element, just a heavier or lighter version of it.
- Mass number A — the total number of nucleons, so A = Z + N. This is the number that matters most in this chapter, because nuclear size and binding energy per nucleon are both organised by A.
The notation is written as AZX: mass number on top, atomic number below, element symbol beside. So 23592U means uranium with 92 protons and 235 − 92 = 143 neutrons. Strictly speaking the Z is redundant — the symbol U already tells you Z = 92 — which is why you will also see the same nucleus written as U-235 or uranium-235. All three mean the same thing, and CBSE uses all three.
One number worth carrying in your head: a proton and a neutron have almost the same mass, about 1.007 u and 1.009 u respectively, while an electron is about 1836 times lighter than a proton. That is why we say the mass of an atom lives almost entirely in its nucleus, and why A — a pure counting number — is such a good stand-in for the mass of a nucleus in unit u.
Aluminium-27: Z = 13 so there are 13 protons; N = A − Z = 27 − 13 = 14 neutrons; nucleons = A = 27; charge = +13e = 13 × 1.6 × 10−19 = 2.08 × 10−18 C.
Uranium-238: 92 protons, N = 238 − 92 = 146 neutrons, A = 238 nucleons, charge = +92e = 1.47 × 10−17 C.
Notice that uranium has far more neutrons than protons (146 versus 92) while aluminium is nearly balanced (14 versus 13). That drift towards extra neutrons in heavy nuclei is not a coincidence — it is the fingerprint of proton-proton electrical repulsion, and it will come back when we read the curve.
Take each nucleon as roughly 1 u = 1.66 × 10−27 kg.
m ≈ A × 1 u = 56 × 1.66 × 10−27 = 9.3 × 10−26 kg.
The true mass of the iron-56 atom is 55.934937 u, so our A-based estimate of 56 u is high by only 0.12 per cent. That is why examiners are happy for you to write “mass of nucleus ≈ A u” in an estimation question — but never in a mass-defect question, where that 0.12 per cent is the entire answer.
Why it works. The whole chapter is a story about a small discrepancy. A is the honest count of nucleons; the measured mass is always slightly less than A u. Everything interesting in Nuclei — binding energy, stability, fission, fusion, the energy of the Sun — hides inside that “slightly less”. So it is worth training your eye now: whenever you see a nuclear mass quoted to six decimal places, the examiner is not showing off. Those decimals are the physics.
The Atomic Mass Unit u and Why 1 u = 931.5 MeV Matters
Kilograms are a terrible unit for nuclei. A proton is 1.67 × 10−27 kg, and if you try to do a mass-defect subtraction in kilograms you will be subtracting two numbers that agree to seven decimal places. So physicists chose a unit that is the right size for the job.
The atomic mass unit (symbol u, sometimes written amu) is defined as exactly one twelfth of the mass of one neutral carbon-12 atom:
and, through E = mc², the same amount of mass is worth
1 u ≡ 931.5 MeV of energy (931.494 MeV if you want the full precision).
Because 12C has 12 nucleons and its mass is exactly 12 u by definition, 1 u is very close to the mass of a single nucleon — which is exactly why the unit is convenient.
Where does 931.5 come from? Nowhere mysterious — it is one multiplication that you should do once, by hand, so that the number stops feeling like a magic spell:
Step 1 — energy in joules.
E = mc² = (1.660539 × 10−27 kg) × (2.99792458 × 108 m/s)²
E = 1.660539 × 10−27 × 8.98755 × 1016 = 1.4924 × 10−10 J
Step 2 — convert joules to electronvolts. Divide by the charge on the electron, because 1 eV = 1.602 × 10−19 J:
E = (1.4924 × 10−10) / (1.602177 × 10−19) = 9.3149 × 108 eV
E = 931.49 MeV ≈ 931.5 MeV. Verified.
Keep the joule value too: 1 u ≡ 1.4924 × 10−10 J. Some questions ask for the answer in joules, and then you multiply your MeV answer by 1.602 × 10−13 J/MeV.
E = 0.02 × 931.5 = 18.63 MeV = 18.63 × 1.602 × 10−13 = 2.985 × 10−12 J.
(b) A nuclear process releases 4.5 MeV. How much mass disappeared?
Δm = 4.5 / 931.5 = 4.83 × 10−3 u, which is 4.83 × 10−3 × 1.66 × 10−27 = 8.02 × 10−30 kg.
That second answer is worth staring at. Eight thousandths of a millionth of a millionth of a millionth of a millionth of a gram, and it is enough energy to knock an electron clean out of every atom it meets.
Making 931.5 MeV feel real: the “mass is missing and that IS the energy” thread. Here is the comparison I want you to carry for the rest of the chapter. In chemistry, when atoms rearrange, the energy per bond is a few electronvolts — when you burn carbon, you get roughly 4 eV per atom. You met that world in carbon and its compounds, where energy is stored in how atoms are arranged. In a nucleus, the energy per nucleon is around 8 million electronvolts. That is a factor of nearly two million, from the same mass of material, and the reason is that chemistry rearranges electrons while nuclear physics rearranges mass itself.
So when a question says “mass defect = 0.030377 u”, do not read it as a small number. Read it as: this much matter stopped existing as matter, and reappeared as 28.30 MeV. The bookkeeping is brutal and exact. Mass in, energy out, at the fixed exchange rate of 931.5 MeV per u.
Size of the Nucleus: R = R₀A1/3 and Why Nuclear Density Ignores A
Scattering experiments — firing fast particles at nuclei and watching how they bounce — give us a way to measure how big a nucleus is. When you collect the results for dozens of different nuclei and plot them, one clean empirical law emerges:
Read it as: the radius grows as the cube root of the number of nucleons. Eight times the nucleons gives only twice the radius. The 1.2 fm is not a nucleon radius — it is an experimental constant that comes out of the fit.
Why a cube root? Because volume, not radius, is what scales with the number of nucleons. Cube both sides and the formula says R³ ∝ A, so the volume of the nucleus (which is (4/3)πR³) is directly proportional to A. Each extra nucleon adds the same fixed amount of volume, exactly the way each extra marble you drop into a bag adds the same amount of bulk. The marbles do not squeeze into each other and they do not spread out either — they just pack.
271/3 = 3 exactly, so R = 1.2 × 3 = 3.6 fm = 3.6 × 10−15 m.
(b) What is the ratio of the radius of a 125Te nucleus to that of 27Al?
R₂/R₁ = (A₂/A₁)1/3 = (125/27)1/3 = 5/3 = 1.667.
Notice the trick in part (b): R₀ cancels, so ratio questions never need the value of R₀ at all. Examiners love choosing A values that are perfect cubes — 8, 27, 64, 125, 216 — so look for them before reaching for a calculator.
Some radii worth having a feel for, all with R₀ = 1.2 fm: A = 27 gives 3.6 fm, A = 64 gives 4.8 fm, A = 125 gives 6.0 fm, A = 216 gives 7.2 fm, and uranium-235 gives 7.4 fm. Notice how slowly it grows — from aluminium to uranium, A goes up nearly nine times but the radius only doubles.
Now for the consequence that CBSE genuinely likes to ask about, and that most students get to only halfway. Take that formula and work out the density of nuclear matter.
Mass: m ≈ A × 1 u = A × 1.66054 × 10−27 kg.
Volume: V = (4/3)πR³ = (4/3)π(R₀A1/3)³ = (4/3)πR₀³A.
Density:
ρ = m/V = [A × 1.66054 × 10−27] / [(4/3)πR₀³A]
The A cancels top and bottom — that is the whole point — leaving
ρ = (1.66054 × 10−27) / [(4/3)π(1.2 × 10−15)³]
(4/3)π(1.2 × 10−15)³ = 7.238 × 10−45 m³
ρ = 2.29 × 1017 kg m−3, the same for every nucleus.
Check it against a real nucleus rather than a formula: iron-56 has R = 1.2 × 561/3 = 4.591 fm and mass 56 u, and dividing gives 2.29 × 1017 kg m−3 again. The cancellation is real, not an approximation artefact.
Why it works. Constant density is a physical statement, not an algebraic accident. It says nuclear matter is incompressible — you cannot squash more nucleons into the same space. Whether you take the lightest nucleus or the heaviest, the nucleons sit at the same spacing, like eggs in an egg tray. This is your first hint about the nuclear force: it must have a strongly repulsive core at very short range, otherwise nucleons would collapse together and heavy nuclei would be denser than light ones. Hold on to that; we will meet it properly in the next section.
How Small Is Small? The Nucleus-and-Atom Scale Gap
Numbers like 10−15 m slide off the mind. So let us make the gap between the atom and its nucleus physical, because this comparison answers a surprising number of one-mark questions.
An atom is about 10−10 m across (that is one angstrom — the size you met for the Bohr orbit of hydrogen). A medium nucleus is about 5 × 10−15 m across. The ratio is roughly 104 to 105 in length, and since volume goes as the cube of length, the ratio of volumes is around 1012 to 1015.
Linear ratio: (1.0 × 10−10)/(4.59 × 10−15) = 2.2 × 104, so the atom is about twenty thousand times wider than its nucleus.
Volume ratio: (2.2 × 104)³ ≈ 1.0 × 1013.
Scale it up so you can see it. If the nucleus were a 1 cm marble, the atom would be 220 m across — two full-length cricket pitches end to end, with one marble at the centre and nothing else but electron cloud. And that marble carries 99.9 per cent of the mass.
One more comparison, using the density we computed: a single teaspoon (5 mL) of pure nuclear matter would have a mass of 2.29 × 1017 × 5 × 10−6 = 1.1 × 1012 kg, about a billion tonnes. That is the density of a neutron star, and it is the same number we computed from R = R₀A1/3.
Why it works. Two different forces set the two sizes. The size of the atom is set by the electrical attraction between the nucleus and light, fast electrons — the same Coulomb physics you worked with in electric charges and fields. The size of the nucleus is set by the nuclear force, which is enormously stronger but reaches only a few femtometres. Strong-but-short pulls things into a tight ball; weak-but-long-range holds them at arm’s length. Hence the factor of ten thousand.
Nuclear Force: Every Feature CBSE Asks You to List
Here is the problem that forces a new force into existence. Inside a uranium nucleus, 92 positive protons sit within about 7 fm of one another. Coulomb’s law says they should repel each other violently — and they do. So why does the nucleus not fly apart in an instant? Gravity is far too feeble to help (it is roughly 1036 times weaker than the electrical repulsion at that scale). Something else must be holding the thing together, and it must be stronger than electrical repulsion at nuclear distances. That something is the nuclear force, also called the strong nuclear force.
CBSE asks about its characteristic features. Learn these five as a list, because that is exactly how the marks are awarded.
2. Extremely short range. It is effective only up to a few femtometres. Beyond about 2.5–3 fm it drops to essentially nothing — two nucleons 10 fm apart do not feel it at all.
3. Charge independent. The force between two protons, between two neutrons, and between a proton and a neutron is the same. The nuclear force does not care about electric charge; only the Coulomb force does.
4. Saturation. Because of the short range, each nucleon interacts only with its immediate neighbours, not with all the others. Add more nucleons far away and nothing extra happens. This is why binding energy per nucleon becomes roughly constant in the middle of the periodic table.
5. Attractive at normal separations, strongly repulsive at very small separations. Beyond about 0.8 fm the nucleons pull each other in; squeeze them closer than that and they push back hard. This repulsive core is what makes nuclear matter incompressible — and it is why nuclear density came out the same for every A.
The potential-energy-versus-separation picture. Draw a horizontal axis for the separation r between two nucleons, and a vertical axis for their mutual potential energy U. Now trace the shape in your mind:
- At large r (beyond a few fm), U is essentially zero — the curve hugs the axis. No force, no energy.
- As r shrinks, U goes negative and dives down. Negative potential energy means attraction and means bound: you would have to supply energy to pull the nucleons apart.
- There is a minimum at r₀ ≈ 0.8 fm, roughly 100 MeV deep for a nucleon pair. This is the comfortable spacing — the bottom of the valley.
- For r < r₀ the curve climbs steeply and crosses zero into large positive values. Positive and rising means violent repulsion. You cannot push nucleons on top of one another.
The everyday analogy that actually fits: two strong magnets with a stiff spring between them. Far apart, nothing happens. Bring them within range and they snap together (attraction, energy released, U goes negative). Try to force them closer than the spring allows and it fights back harder and harder (the repulsive core). The “valley floor” is where they naturally sit.
U = (1/4πε₀)(e²/r), and it is worth remembering that (1/4πε₀)e² = 1.44 MeV·fm. So in nuclear units the whole calculation is a one-liner:
U = 1.44/2.0 = 0.72 MeV (that is 1.15 × 10−13 J).
Compare: binding energy per nucleon in a mid-weight nucleus is about 8.5 MeV. So at a separation of 2 fm the nuclear attraction is more than ten times stronger than the electrical repulsion — the nucleus is comfortably bound.
But notice what happens as Z grows. Coulomb repulsion is long range, so every proton repels every other proton: the total repulsion climbs roughly as Z². The nuclear force saturates and only climbs as A. Sooner or later the long-range effect catches up. That crossover is the reason the binding energy per nucleon curve turns over and starts falling.
Mass-Energy Relation E = mc² and Mass Defect
Einstein’s 1905 result is the hinge of this chapter, so let us state it carefully. Mass and energy are not two separate things that can be converted into each other by some mechanism. They are the same thing, measured in different units, and the exchange rate is c²:
Because c² = 9 × 1016 in SI units, a truly tiny mass carries a colossal energy. One kilogram of anything, fully converted, would give 9 × 1016 J. That is why nuclear processes can look like they break energy conservation — they do not; some of the mass simply stopped being mass.
Now the measurement that changed everything. Weigh the pieces of a nucleus separately, add them up, and then weigh the assembled nucleus. The assembled nucleus is always lighter. That shortfall is the mass defect:
and if you are working with atomic masses (which is what question papers almost always give you), use the atomic mass of hydrogen in place of the proton mass:
Δm = [Z mH + N mn] − Matom
Both versions give the same Δm, because the Z electrons you sneaked in with Z hydrogen atoms are exactly the Z electrons already inside the neutral atom on the right. They cancel. What you must never do is mix the two — hydrogen atomic mass with nuclear mass, or proton mass with atomic mass.
Standard values you will be given, and should recognise on sight: atomic mass of 1H = 1.007825 u, mass of a neutron = 1.008665 u, mass of a proton = 1.007276 u, mass of an electron = 0.000549 u. Notice mH = mp + me to six decimal places, which is the cancellation above, made visible.
Step 1 — add the parts. Z = 2, N = 2:
2 × 1.007825 = 2.015650 u
2 × 1.008665 = 2.017330 u
Total mass of parts = 4.032980 u
Step 2 — subtract the whole.
Δm = 4.032980 − 4.002603 = 0.030377 u
That is it. Two lines. The helium nucleus is 0.030377 u lighter than the four particles it is made of — about 0.75 per cent of its own mass has gone missing, permanently, and turned into the energy that holds it together. Follow the sticky-note flow above: this same Δm becomes 28.30 MeV in the very next section.
Why it works. Think about what you would have to do to pull that helium nucleus apart. You would have to fight the nuclear force all the way out of its potential-energy valley — and that costs work. Energy you put in has to go somewhere, and where it goes is into mass: the separated nucleons weigh more than the bound nucleus by exactly the energy you supplied, divided by c². Run the film backwards and it is the same statement: when free nucleons snap together, they release energy, and the mass books balance by the nucleus coming out lighter. A bound system is always lighter than its free parts. That is not special to nuclei — it is true for atoms and for planets too — but only in nuclei is the effect big enough to weigh.
Z = 1, N = 1, so parts = 1.007825 + 1.008665 = 2.016490 u
Δm = 2.016490 − 2.014102 = 0.002388 u
Energy = 0.002388 × 931.5 = 2.224 MeV
So a gamma ray of at least 2.224 MeV can break a deuteron apart, and a photon of 2.0 MeV cannot, no matter how many of them you send. The deuteron is the most loosely bound nucleus that exists — only 1.11 MeV per nucleon, against 8.79 MeV per nucleon for iron. Remember this number; it is why deuterium is such useful fuel for fusion.
Binding Energy and Binding Energy Per Nucleon
Convert the mass defect into energy and you have the binding energy of the nucleus:
B.E. per nucleon = B.E. / A
Binding energy is the energy you must supply to break the nucleus into free, separated nucleons — equivalently, the energy released if you could assemble it from free nucleons. Binding energy per nucleon is that total shared out over the nucleons, and it is the true measure of stability.
Why divide by A? Because total binding energy is a misleading way to compare nuclei. Uranium-235 has a total binding energy of 1783.9 MeV and helium-4 only 28.3 MeV, which makes uranium sound sixty times more stable. It is not — uranium simply has more nucleons to bind. Per nucleon, uranium manages 7.59 MeV while iron manages 8.79 MeV, and it is iron that is the more stable arrangement. Whenever a question asks “which is more stable?”, it is asking about binding energy per nucleon, never about the total.
Here is the analogy I find lands best. A group buys a bundle deal: the bundle costs less than the sum of the individual prices, and the difference is the discount. Mass defect is the discount. Binding energy is the discount expressed in energy. And binding energy per nucleon is the discount per person — the only fair way to compare a bundle of four with a bundle of 235.
Parts: Z = 8, N = 8
8 × 1.007825 = 8.062600 u
8 × 1.008665 = 8.069320 u
Total = 16.131920 u
Mass defect: Δm = 16.131920 − 15.994915 = 0.137005 u
Binding energy: 0.137005 × 931.5 = 127.62 MeV
Per nucleon: 127.62 / 16 = 7.976 MeV
Sanity check before you write the final answer: is your per-nucleon value between 1 and 9 MeV? Every nucleus in nature is. If you get 0.13 MeV or 480 MeV per nucleon, you have divided by the wrong thing or forgotten the 931.5.
Iron-56 (Z = 26, N = 30):
parts = 26(1.007825) + 30(1.008665) = 26.203450 + 30.259950 = 56.463400 u
Δm = 56.463400 − 55.934937 = 0.528463 u
B.E. = 0.528463 × 931.5 = 492.26 MeV, so B.E./A = 492.26/56 = 8.790 MeV
Uranium-235 (Z = 92, N = 143):
parts = 92(1.007825) + 143(1.008665) = 92.719900 + 144.239095 = 236.958995 u
Δm = 236.958995 − 235.043930 = 1.915065 u
B.E. = 1.915065 × 931.5 = 1783.88 MeV, so B.E./A = 1783.88/235 = 7.591 MeV
Conclusion: uranium has 3.6 times the total binding energy, yet iron is the more tightly bound and more stable nucleus, because 8.790 > 7.591 MeV per nucleon. That gap of 1.2 MeV per nucleon, multiplied by 235 nucleons, is precisely the energy available in nuclear fission — which is where we are heading.
The process is 4He → 3H + 1H.
Mass of products = 3.016049 + 1.007825 = 4.023874 u
Mass of 4He = 4.002603 u
Increase in mass = 4.023874 − 4.002603 = 0.021271 u
Energy required = 0.021271 × 931.5 = 19.81 MeV
Compare that with the total binding energy of helium-4, 28.30 MeV, and with the average per nucleon, 7.07 MeV. All three numbers are different and all three are correct answers to different questions. Total B.E. dismantles the nucleus completely; 19.81 MeV removes one specific proton and leaves a bound 3H behind; 7.07 MeV is only an average. Read the question’s verb: “completely separate” means total, “remove a nucleon” means a separation energy like this one.
These are all the same three lines of arithmetic. Here is the table of computed values we will be using for the rest of the page — every row was produced by the method of Example 11, from the standard atomic masses in u:
| Nuclide | A | Mass defect Δm (u) | Binding energy (MeV) | B.E. per nucleon (MeV) |
|---|---|---|---|---|
| 2H | 2 | 0.002388 | 2.224 | 1.112 |
| 4He | 4 | 0.030377 | 28.296 | 7.074 |
| 12C | 12 | 0.098940 | 92.163 | 7.680 |
| 16O | 16 | 0.137005 | 127.620 | 7.976 |
| 56Fe | 56 | 0.528463 | 492.263 | 8.790 |
| 120Sn | 120 | 1.095598 | 1020.550 | 8.505 |
| 208Pb | 208 | 1.756788 | 1636.448 | 7.868 |
| 235U | 235 | 1.915065 | 1783.883 | 7.591 |
| 238U | 238 | 1.934202 | 1801.709 | 7.570 |
Binding Energy Per Nucleon Curve Explained: Every Conclusion in One Graph
This is the graph the whole chapter hangs on, so take a minute with it. Along the bottom is the mass number A. Up the side is binding energy per nucleon in MeV. Every white dot is a real nuclide whose value we computed from its measured atomic mass by the method of Example 11 — nothing here is sketched by eye.
Now read it in four movements, left to right.
- A steep climb (A = 1 to about 20). Hydrogen-1 has no binding energy at all — a single nucleon has nothing to be bound to. The deuteron manages a feeble 1.11 MeV per nucleon. Then helium-4 leaps to 7.07, and by oxygen-16 we are at 7.98. The rise is dramatic and jagged.
- A long, almost flat plateau (A ≈ 20 to 120). Values sit between about 8.0 and 8.8 MeV per nucleon — nearly constant. This flatness is saturation in action: a nucleon deep inside a nucleus is surrounded by as many neighbours as the short-range force can reach, so adding more nucleons far away changes nothing for it.
- A broad, gentle maximum around A ≈ 56 to 62. Iron-56 sits at 8.790 MeV per nucleon; its neighbour nickel-62 is fractionally higher at 8.795. For exam purposes the peak is “around A = 56, iron”, and its height is “about 8.8 MeV per nucleon”. These are the most tightly bound nuclei in the universe.
- A slow decline (A > 62 all the way to uranium). Tin-120 is at 8.505, lead-208 at 7.868, uranium-235 at 7.591. Nothing dramatic — just a steady sag of about 1.2 MeV per nucleon over 180 nucleons.
Now the important part: what each feature means. This is the section examiners mine for 3-mark questions, and almost every conclusion is a sentence you can read straight off the shape.
2. Mid-weight nuclei are the most stable. Highest B.E./A means most tightly bound. Around A = 56 nature is at its most comfortable.
3. The fall at large A is the Coulomb tax. Proton-proton repulsion is long range, so it grows roughly as Z² while nuclear binding only grows as A. In heavy nuclei that tax eats into the binding, so B.E./A sags — and the nucleus becomes vulnerable to splitting.
4. The steep left-hand rise means light nuclei gain hugely by joining. Two light nuclei sitting at 1–3 MeV per nucleon can move to 7 MeV per nucleon by fusing. That is a big climb, so a lot of energy comes out.
5. Heavy nuclei gain by splitting. A nucleus at 7.6 MeV per nucleon breaking into fragments near 8.5 MeV per nucleon climbs 0.9 MeV per nucleon — and with 235 nucleons in the queue, that is around 200 MeV.
6. Some nuclei stick out above the trend. Helium-4 at 7.07 MeV per nucleon sits far above its neighbours (deuterium 1.11, tritium 2.83, lithium-6 5.33). Nuclei with even, matched numbers of protons and neutrons — He-4, C-12, O-16 — are unusually tightly bound. That is why alpha particles exist as intact objects and why helium is the ash of stars.
(a) Which is more stable, 7Li (B.E./A = 5.61 MeV) or 4He (7.07 MeV)? Helium-4, because it has the higher binding energy per nucleon — even though lithium-7 has the larger total binding energy (39.25 MeV against 28.30 MeV).
(b) Will energy be released if two 2H nuclei combine to form 4He? Yes. You are moving nucleons from 1.11 MeV per nucleon up to 7.07 MeV per nucleon — uphill on the curve. Rough size of the release: 4 nucleons × (7.07 − 1.11) = 23.8 MeV, and the exact mass calculation gives 23.85 MeV. The curve gets you within a fraction of a per cent.
(c) Will energy be released if 56Fe splits into two 28Si nuclei? No. Iron-56 is at 8.790 and silicon-28 is at 8.448 MeV per nucleon — that is downhill, so you would have to supply about 56 × (8.790 − 8.448) = 19 MeV to make it happen. Iron does not want to split and it does not want to fuse. It is the bottom of the barrel, the end of the line.
Why it works. Part (b) of that example is the key move of the whole chapter, so let us be explicit about the logic. The total binding energy of a collection of nucleons is A × (B.E./A). If a rearrangement leaves the nucleons with a higher B.E./A, the total binding energy has gone up — and total binding energy is nothing but stored mass that has been converted away. More binding means less mass means energy handed out. So the single question you ask of any nuclear reaction is: did the nucleons end up higher on the curve than they started? If yes, energy comes out.
The Iron Hill Rule: One Memory Device for the Whole Chapter
Students lose marks in this chapter for one silly reason: under pressure they cannot remember which way round the energy goes. Does fission release or absorb? Does fusion? Which end of the periodic table does what? So here is a device I want you to own. I call it the Iron Hill Rule, and it is three words long.
Picture the binding energy per nucleon curve as a hill with iron at the summit. Every nucleus in the universe is standing somewhere on that hill, and every nucleus would rather be higher up.
• Light nuclei are on the steep left slope. To climb, they must join together — that is fusion.
• Heavy nuclei are on the gentle right slope. To climb, they must break apart — that is fission.
• Iron is at the top. It can do neither, so it does nothing. It is the ash at the end of every stellar fire.
Whichever direction the nucleons travel, climbing the hill releases energy. Going downhill would cost you energy, which is why you never see it happen spontaneously.
Make it physical with your hand. Hold your palm in front of you sloping steeply up from the left and gently down to the right — that is the curve. Tap the far left: “hydrogen, must join.” Tap the far right: “uranium, must split.” Tap the top: “iron, stuck.” Do that three times now and you will still have it in the exam hall in March.
And here is the bonus: the rule also tells you the size of the release without any masses at all. Energy released ≈ (number of nucleons) × (climb in MeV per nucleon). A short climb with many nucleons gives fission’s 200 MeV. A tall climb with few nucleons gives fusion’s 17.6 MeV — smaller in total, but far bigger per nucleon, which is why fusion is the better deal per kilogram of fuel.
Nuclear Fission: Where the 200 MeV Actually Comes From
Nuclear fission is the breaking of a heavy nucleus into two lighter nuclei of comparable size, with a release of energy. It is not the same as a nucleus shedding a small chunk; the defining feature is that the nucleus splits into two big pieces.
The classic case is uranium-235 absorbing a slow neutron. The neutron makes the nucleus wobble and stretch until electrical repulsion between the two ends wins over the short-range nuclear attraction holding the middle together, and the nucleus snaps in two. A typical outcome is
Check the books, as an examiner would: mass numbers, 235 + 1 = 236 on the left, 141 + 92 + 3 = 236 on the right. Charges, 92 + 0 = 92 on the left, 56 + 36 + 0 = 92 on the right. Both balance. That check is worth a mark on its own and it costs you five seconds.
Fission is not tidy. The fragments are not always barium and krypton — the split is usually asymmetric, and dozens of fragment pairs are possible, releasing on average two to three neutrons. What is remarkably steady is the energy: about 200 MeV per fission, almost all of it appearing as kinetic energy of the two fragments flying apart.
Now the point of this whole page: where does the 200 MeV come from? Not from the neutron, which brings in a fraction of an eV. It comes from the climb up the Iron Hill. Uranium-235 sits at 7.591 MeV per nucleon. Fragments with A around 100 to 140 sit near 8.4 to 8.5 MeV per nucleon. Every one of the 235 nucleons gets promoted by roughly 0.9 MeV, and 235 × 0.9 is about 200. That is the entire explanation.
Method: energy released = (total binding energy after) − (total binding energy before). More binding after means mass was lost and energy came out.
Before: B.E. = 240 × 7.6 = 1824 MeV
After: B.E. = 2 × (120 × 8.5) = 2 × 1020 = 2040 MeV
Energy released = 2040 − 1824 = 216 MeV
Same thing in one line with the Iron Hill Rule: 240 nucleons × (8.5 − 7.6) MeV climbed = 216 MeV. Identical answer, one third of the writing.
Watch the sign convention. Because binding energy is energy you would have to put in, an increase in total binding energy means energy has been given out. If your answer comes out negative, the reaction absorbs energy instead — and it will not happen spontaneously.
(a) Count the nuclei. One mole of 235U is 235 g and contains 6.022 × 1023 nuclei, so
N = 6.022 × 1023 / 235 = 2.563 × 1021 nuclei per gram
Energy = 2.563 × 1021 × 200 MeV = 5.13 × 1023 MeV
In joules: 5.13 × 1023 × 1.602 × 10−13 = 8.21 × 1010 J (that is 2.28 × 104 kWh)
(b) The missing mass. Δm = E/c² = 8.21 × 1010 / 9 × 1016 = 9.1 × 10−7 kg = 0.91 mg. Out of a full gram, less than a milligram — about one part in a thousand — actually stopped being matter.
(c) Against coal. Burning one gram of coal gives roughly 30 kJ. The ratio is 8.21 × 1010 / 3 × 104 ≈ 2.7 million. Per atom the gap is even starker: 200 MeV for one fission against a few eV per atom in a chemical fire.
The counting step in part (a) is pure mole arithmetic — if Avogadro’s number and moles feel shaky, ten minutes on moles and concentration in Solutions will pay for itself here.
One consequence worth naming, since every fission releases two or three fresh neutrons: those neutrons can trigger further fissions, which release more neutrons, and so on. That is a chain reaction. Controlling it is the business of a nuclear reactor — and the reactor itself is not part of your 2026-27 syllabus, so it lives in the clearly-labelled amber section further down rather than here.
Nuclear Fusion, the Coulomb Barrier and Why Stars Shine
Nuclear fusion is the joining of two light nuclei into a heavier one, with a release of energy. On the Iron Hill it is the mirror image of fission: same hill, same rule, opposite end, opposite direction of travel.
Three reactions do the heavy lifting in every textbook, and all three are worth knowing by heart:
21H + 21H → 31H + 11H + 4.03 MeV
21H + 31H → 42He + 10n + 17.59 MeV
Now the obvious question: if fusion is so profitable, why is the world not running on it? Because of one very stubborn obstacle.
That wall of electrical potential energy is the Coulomb barrier, and it is roughly 0.5 to 1 MeV high for the lightest nuclei. To fuse, the nuclei must arrive with enough kinetic energy to climb it — which means enormous speed, which means enormous temperature, of the order of 109 K and above. This is why fusion is called a thermonuclear process.
Branch 1: 2H + 2H → 3He + n
Mass before = 2 × 2.014102 = 4.028204 u
Mass after = 3.016029 + 1.008665 = 4.024694 u
Δm = 0.003510 u → E = 0.003510 × 931.5 = 3.27 MeV
Branch 2: 2H + 2H → 3H + 1H
Mass after = 3.016049 + 1.007825 = 4.023874 u
Δm = 4.028204 − 4.023874 = 0.004330 u → E = 4.03 MeV
A note on electron bookkeeping, because it matters: both sides of each equation have the same total number of protons (2), so if you use atomic masses throughout, the electrons cancel exactly and no correction is needed. Consistency is everything — atomic masses on both sides, or nuclear masses on both sides, never a mixture.
Mass before = 2.014102 + 3.016049 = 5.030151 u
Mass after = 4.002603 + 1.008665 = 5.011268 u
Δm = 0.018883 u
E = 0.018883 × 931.5 = 17.59 MeV
Per nucleon: five nucleons take part, so 17.59/5 = 3.52 MeV per nucleon.
Fission for comparison: 200 MeV shared over 235 nucleons = 0.85 MeV per nucleon.
So fission wins per event (200 MeV against 17.6 MeV) but fusion wins per nucleon by more than four times — and it is nucleons you have to buy and carry. That, plus the fact that its fuel is hydrogen rather than mined uranium and its products are not long-lived radioactive waste, is why fusion is the prize.
Cross-check straight off the curve, with no masses at all: the total binding energy before the reaction is 2.224 MeV (deuterium) + 8.482 MeV (tritium) = 10.706 MeV; afterwards the helium-4 nucleus holds 28.296 MeV and the free neutron holds none. The gain in binding energy is 28.296 − 10.706 = 17.59 MeV — exactly the mass-defect answer, obtained purely by climbing the hill.
(a) U = (1/4πε₀)(e²/r) = 1.44/2.0 = 0.72 MeV = 1.15 × 10−13 J
(b) Average thermal kinetic energy is (3/2)kBT, so
T = 2U/(3kB) = (2 × 1.15 × 10−13)/(3 × 1.38 × 10−23) = 5.6 × 109 K
Nearly six billion kelvin. And here is the puzzle every good student spots: the core of the Sun is only about 1.5 × 107 K — four hundred times too cold. So how does the Sun run at all?
Two reasons, and CBSE is happy with them stated qualitatively. First, temperature gives an average; the Maxwell distribution has a high-energy tail, and the rare fastest protons far exceed the average. Second, quantum mechanics allows a particle to tunnel through a barrier it cannot classically climb. The Sun runs on its luckiest, fastest protons plus tunnelling — which is exactly why it burns so slowly and has lasted 4.6 billion years instead of exploding.
Stellar energy: the proton-proton cycle, qualitatively. The Sun is a ball of hydrogen at about 1.5 × 107 K in its core, and its output comes from turning hydrogen into helium in a sequence of small steps rather than one grand collision (four protons meeting at the same instant is hopelessly unlikely). The chain, with the energy released at each step, runs like this:
e+ + e− → γ + γ (1.02 MeV, the positron annihilating with an electron)
21H + 11H → 32He + γ (5.49 MeV)
32He + 32He → 42He + 11H + 11H (12.86 MeV)
The first three steps have to happen twice to feed the fourth, so the net effect is: four protons go in, one helium-4 nucleus comes out, and the books balance at about 26.7 MeV released per helium nucleus formed. In heavier, hotter stars a different route with carbon as a catalyst (the carbon-nitrogen-oxygen cycle) does the same job. Either way, it is fusion, and either way the products are climbing the Iron Hill.
Cross-check with masses: four hydrogen atoms weigh 4 × 1.007825 = 4.031300 u, while a helium-4 atom weighs 4.002603 u. Δm = 0.028697 u, so E = 0.028697 × 931.5 = 26.73 MeV. Two independent routes, same answer.
(b) How much mass does the Sun lose each second? Its power output is 3.85 × 1026 W, so
Δm/Δt = P/c² = 3.85 × 1026 / 9.0 × 1016 = 4.3 × 109 kg per second
That is about 4.3 million tonnes of the Sun’s mass turning into sunlight every second — and it has been doing this for 4.6 billion years, having spent well under a thousandth of itself. Note also that only about 0.7 per cent of the mass of the hydrogen consumed is converted (26.73 MeV out of 4 × 931.5 × 1.007825 MeV of rest mass), which is why stars last so long.
Why Fission AND Fusion Both Release Energy — The One Unifying Idea
This is the question that separates students who have understood the chapter from students who have memorised it. Fission breaks a big nucleus into small ones. Fusion joins small nuclei into a bigger one. They are exact opposites. How can both give out energy? Surely one of them must cost energy?
The trap in that question is the assumption that “big” and “small” is what matters. It is not. What matters is where the nucleons end up on the binding energy per nucleon curve — and the curve is not a straight line. It is a hill with a summit near iron. Both ends of the periodic table are below that summit, so both ends can climb, and they climb by moving in opposite geographical directions.
Fission does this by going down in mass number from 235 towards 120. Fusion does this by going up in mass number from 2 towards 4. Different directions along the horizontal axis; the same direction — upwards — along the vertical one. That is the whole answer, and it is worth every mark it carries.
Let us do the arithmetic side by side one final time, using nothing but the curve.
| Question | Fission of 235U | Fusion of 2H + 3H |
|---|---|---|
| Where do the nucleons start on the curve? | 7.59 MeV per nucleon | 1.11 and 2.83 MeV per nucleon |
| Where do they finish? | about 8.5 MeV per nucleon | 7.07 MeV per nucleon |
| Direction along the A axis | A decreases (235 → ~120) | A increases (2, 3 → 4) |
| Direction on the curve | Uphill | Uphill |
| So energy is… | released, about 200 MeV | released, 17.59 MeV |
And now the two bonus questions that the same picture answers for free.
Why is iron the end of the line? Because it is at the summit. An iron nucleus cannot climb by splitting (its fragments would sit lower) and cannot climb by fusing (heavier nuclei sit lower too). Every path from iron is downhill, and downhill costs energy instead of yielding it. This is not just an exam fact — it is why a massive star that has fused its way up to an iron core has no fuel left and collapses. The Iron Hill has a summit, and stars die on it.
Why is helium-4 unusually stable? Look at its dot on the curve: 7.07 MeV per nucleon, sitting far above its neighbours deuterium (1.11), tritium (2.83) and lithium-6 (5.33). Helium-4 has two protons and two neutrons — an even, matched, closed little set — and such configurations bind exceptionally well. Carbon-12 and oxygen-16 show the same bulge. This is why the α particle exists as an intact object that heavy nuclei can emit whole, and why helium is the natural ash of hydrogen burning in stars.
Nuclear Fission and Fusion Difference Class 12: Side-by-Side Table
This comparison is a frequent 2- or 3-mark question, and the marks go to differences, not descriptions. Learn the rows, not the paragraphs.
| Point of comparison | Nuclear fission | Nuclear fusion |
|---|---|---|
| What happens | A heavy nucleus splits into two nuclei of comparable size | Two light nuclei join to form a heavier nucleus |
| Typical fuel | 235U, 239Pu (mass numbers above ~230) | Isotopes of hydrogen: 2H, 3H |
| Trigger | A slow (thermal) neutron — uncharged, so no barrier to overcome | Extreme temperature to beat the Coulomb barrier |
| Conditions needed | Works at ordinary temperature | Needs of the order of 107–109 K |
| Energy per event | Large: about 200 MeV per fission | Smaller: 3–18 MeV per reaction |
| Energy per nucleon | About 0.85 MeV per nucleon | About 3.5 MeV per nucleon — four times better |
| Products | Unstable fragments plus 2–3 free neutrons | Mostly stable light nuclei such as 4He |
| Chain reaction possible? | Yes — the released neutrons can trigger more fissions | No chain mechanism of this kind |
| Where it happens naturally | Rare in nature; engineered in reactors and weapons | The power source of the Sun and every star |
| Reason energy is released | Identical for both: the products have a higher binding energy per nucleon, so mass has been converted into energy | |
Not in Your 2026-27 Exam: Radioactivity, Decay Law, Half-Life, Reactor
Everything in this section — radioactivity, α, β and γ decay, the radioactive decay law, half-life, mean life, activity, and the nuclear reactor — is outside that scope. It is here because your textbook, your coaching notes and most websites still carry it, and I would rather you know exactly where the boundary is than lie awake wondering. Read it for understanding, for JEE or NEET, or out of pure interest. Do not spend board-exam revision time on it.
Radioactivity is the spontaneous emission of radiation by unstable nuclei. Discovered by Becquerel in 1896, it comes in three flavours, and the reason it happens at all is the curve you have just learned: nuclei that sit below the summit of the Iron Hill, or that have a badly lopsided proton-to-neutron ratio, can reach a more tightly bound arrangement by throwing something away.
- α decay — the nucleus emits a helium-4 nucleus (2 protons, 2 neutrons). A drops by 4 and Z drops by 2. Heavy nuclei favour this because the α particle is that unusually tightly bound object we noticed on the curve.
- β− decay — a neutron inside the nucleus turns into a proton, emitting an electron and an antineutrino. A stays the same, Z increases by 1. This is how a neutron-rich nucleus rebalances.
- β+ decay — a proton turns into a neutron, emitting a positron and a neutrino. A stays the same, Z decreases by 1. You saw exactly this in the first step of the proton-proton cycle.
- γ decay — the nucleus sheds surplus energy as a high-energy photon. Neither A nor Z changes; only the internal energy does. This is the nuclear analogue of an excited atom emitting light.
The radioactive decay law says the rate of decay is proportional to the number of undecayed nuclei present: dN/dt = −λN, which integrates to N = N₀e−λt, where λ is the decay constant. From it come two quantities students often confuse: the half-life T1/2 = (ln 2)/λ = 0.693/λ, which is the time for half of a sample to decay, and the mean life τ = 1/λ, the average lifetime of a single nucleus. Since T1/2 = 0.693τ, the half-life is always the shorter of the two. A useful shortcut for the non-examinable numericals: after n half-lives, the fraction remaining is (1/2)n.
The nuclear reactor takes the chain reaction from the fission section and tames it. Fuel rods of enriched uranium supply the 235U; a moderator (heavy water or graphite) slows the fast neutrons down, because 235U captures slow neutrons far more readily; control rods of cadmium or boron absorb surplus neutrons so that on average exactly one neutron from each fission goes on to cause the next one; a coolant carries the heat away to boil water and drive a turbine, at which point it becomes the ordinary electric current problem you already know how to solve. The engineering is beautiful, but for the 2026-27 board exam it is background reading.
Nuclei Class 12 Physics Important Questions: What Examiners Actually Ask
The question types in this chapter repeat with remarkable loyalty. Here is the honest shortlist, with what each one is really testing, so you can aim your revision instead of spraying it.
| Question type | Usual marks | What it is really testing |
|---|---|---|
| Find Δm, B.E. and B.E./A from given masses | 2–3 | Careful subtraction and the 931.5 conversion |
| Draw the B.E./A curve and state three conclusions | 3 | Labelled axes, peak near A = 56, fission and fusion read off it |
| Why do both fission and fusion release energy? | 2–3 | The one unifying idea — products higher on the curve |
| Show nuclear density is independent of A | 2 | Showing the A cancel, not the final number |
| List the features of the nuclear force | 2–3 | Short range, charge independence, saturation, repulsive core |
| Energy released in a given fission or fusion reaction | 3 | Balancing A and Z, then mass before minus mass after |
| Why does fusion need high temperature? | 2 | The Coulomb barrier, and why a neutron does not face one |
| Ratio of nuclear radii of two given nuclides | 1–2 | R ∝ A1/3, and spotting that R₀ cancels |
Notice what is not on that list: nothing about decay constants, nothing about reactors, nothing that needs a memorised table of isotopes. Eight question types, all built out of the same three or four moves. When you can do the worksheet below cold, you have covered the chapter. If you want more of this subject, the whole set of chapters lives in the Class 12 Physics notes archive.
Common Mistakes and Unit Traps in Nuclei Numericals
Almost every mark lost in this chapter is lost in one of these seven ways. Read them the night before the exam.
2. Mixing atomic and nuclear masses. Either use mH = 1.007825 u with atomic masses, or mp = 1.007276 u with nuclear masses. Never one of each.
3. Forgetting to divide by A. “Binding energy” and “binding energy per nucleon” are different questions with answers that differ by a factor of up to 235.
4. Getting the direction of energy flow backwards. More total binding energy after the reaction means energy released. Chant it: more binding, less mass, energy out.
5. Using N instead of A, or A instead of N. In Δm = ZmH + Nmn − M, the neutron count is N = A − Z. Writing A mn is an instant wrong answer.
6. Unit slips in the final conversion. 1 u → 931.5 MeV; 1 MeV → 1.602 × 10−13 J; 1 eV → 1.602 × 10−19 J. If a question asks for joules and you stop at MeV, you lose the last mark.
7. Saying “the nuclear force is like a strong electric force”. It is charge independent. Neutrons feel it exactly as protons do.
Practice Worksheet
Twelve questions, arranged as a staircase from one-step substitution to full exam-style energy balances. Do them with a pen and a calculator and without opening the answers first — the whole value is in the struggle before the reveal. Where masses are needed they are given in the question, and every answer below has been computed and checked.
Useful constants: 1 u = 931.5 MeV/c² = 1.66054 × 10−27 kg; mH = 1.007825 u; mn = 1.008665 u; R₀ = 1.2 fm; 1 MeV = 1.602 × 10−13 J; (1/4πε₀)e² = 1.44 MeV·fm; kB = 1.38 × 10−23 J/K; NA = 6.022 × 1023.
Q1. For the nucleus 4020Ca, write down the number of protons and neutrons, and calculate its radius.
R = R₀A1/3 = 1.2 × 401/3 fm. Since 401/3 = 3.420, R = 1.2 × 3.420 = 4.10 fm = 4.10 × 10−15 m.
Note: calcium-40 has equal protons and neutrons, which is typical of stable light and medium nuclei — the neutron surplus only appears in heavy ones.
Q2. Find the ratio of the nuclear radii of 208Pb and 27Al. Does your answer depend on the value of R₀?
No, it does not depend on R₀ at all — R₀ appears in both radii and cancels in the ratio. That is why ratio questions can be answered even if the paper never quotes R₀.
Reality check: lead has nearly eight times as many nucleons as aluminium, yet is only about twice as wide. Cube-root growth is slow.
Q3. The atomic mass of 73Li is 7.016003 u. Find its mass defect, binding energy and binding energy per nucleon.
Mass of parts = 3(1.007825) + 4(1.008665) = 3.023475 + 4.034660 = 7.058135 u
Δm = 7.058135 − 7.016003 = 0.042132 u
B.E. = 0.042132 × 931.5 = 39.25 MeV
B.E./A = 39.25/7 = 5.61 MeV per nucleon
Read it off the curve: 5.61 MeV per nucleon puts lithium-7 well up the steep left-hand slope but far below helium-4 at 7.07 — another reminder that binding energy per nucleon does not rise smoothly with A in the light region.
Q4. How much energy is needed to remove one neutron from a 4He nucleus? Take 4He = 4.002603 u and 3He = 3.016029 u. Why is your answer not 28.30/4 MeV?
Mass of products = 3.016029 + 1.008665 = 4.024694 u
Mass increase = 4.024694 − 4.002603 = 0.022091 u
Energy required = 0.022091 × 931.5 = 20.58 MeV
It is not 28.30/4 = 7.07 MeV because 7.07 MeV is only the average binding energy per nucleon. Pulling one neutron out of helium-4 leaves a 3He nucleus that is very poorly bound (only 2.57 MeV per nucleon), so you have to pay for wrecking the tight four-nucleon arrangement as well as for removing the neutron. Averages describe the whole; separation energies describe one specific removal.
Q5. A nucleus with A = 210 and binding energy per nucleon 7.7 MeV splits into two equal fragments with A = 105 and binding energy per nucleon 8.4 MeV. Find the energy released.
Total B.E. before = 210 × 7.7 = 1617 MeV
Total B.E. after = 2 × (105 × 8.4) = 2 × 882 = 1764 MeV
Energy released = 1764 − 1617 = 147 MeV
Or in one line with the Iron Hill Rule: 210 × (8.4 − 7.7) = 210 × 0.7 = 147 MeV. The products are higher on the curve, so the process is exothermic and can occur spontaneously (given a trigger).
Q6. If every nucleus in 1.0 kg of 235U fissions, releasing 200 MeV each, for how long could the energy run a 100 MW power station at 100 per cent efficiency?
Number of nuclei: N = (1000 g / 235 g mol−1) × 6.022 × 1023 = 2.563 × 1024
Total energy: E = 2.563 × 1024 × 200 × 1.602 × 10−13 J = 8.21 × 1013 J
Time: t = E/P = 8.21 × 1013 / 1.0 × 108 = 8.21 × 105 s
t = 8.21 × 105 / 86400 = 9.5 days
One kilogram. Nine and a half days of a hundred-megawatt station. Do the same sum for coal and you would need roughly 2700 tonnes.
Q7. Three 4He nuclei fuse to form one 12C nucleus. Using 4He = 4.002603 u and 12C = 12.000000 u exactly, find the energy released.
Mass before = 3 × 4.002603 = 12.007809 u
Mass after = 12.000000 u
Δm = 0.007809 u
E = 0.007809 × 931.5 = 7.27 MeV
Check it against the curve: helium-4 is at 7.074 and carbon-12 at 7.680 MeV per nucleon, so the climb is 12 × (7.680 − 7.074) = 7.27 MeV. The two methods agree exactly, as they must.
Why it matters: this is the reaction (the “triple-alpha process”) that manufactures every carbon atom in your body inside old stars. The climb is small, which is why it needs a red giant’s core to get going.
Q8. Prove that nuclear density is independent of mass number, and evaluate it. Compare it with the density of water.
Mass of nucleus, m ≈ A u. Volume, V = (4/3)πR³ = (4/3)πR₀³A.
ρ = m/V = Au / [(4/3)πR₀³A] = u / [(4/3)πR₀³] — the A has cancelled, so ρ is the same for every nucleus.
Substituting: (4/3)π(1.2 × 10−15)³ = 7.238 × 10−45 m³
ρ = (1.66054 × 10−27) / (7.238 × 10−45) = 2.29 × 1017 kg m−3
Water is 1.0 × 103 kg m−3, so nuclear matter is about 2.3 × 1014 times denser than water. The physical meaning: nuclear matter is incompressible, because the nuclear force has a repulsive core at very small separations.
Q9. Why does the binding energy per nucleon curve rise steeply for small A, stay almost flat in the middle, and fall slowly for large A? Give one reason for each region.
Steep rise (small A): in a very small nucleus most nucleons are on the surface with few neighbours to bind to. Each nucleon you add gives the existing ones more partners, so the average binding per nucleon climbs quickly.
Flat middle (A ≈ 20–120): saturation. The nuclear force is short range, so once a nucleon is surrounded by as many neighbours as it can reach, adding nucleons elsewhere in the nucleus makes no difference to it. Binding energy then grows in proportion to A, so B.E./A is constant.
Slow fall (large A): the Coulomb repulsion between protons is long range, so every proton repels every other one and the total repulsion grows roughly as Z², faster than the binding grows with A. This surplus repulsion eats into the binding and B.E./A sags — which is exactly what makes heavy nuclei liable to fission.
Q10. Two protons must come within 3.0 fm of each other to fuse. Find the Coulomb barrier in MeV and estimate the temperature needed, using average kinetic energy = (3/2)kBT.
Barrier: U = (1/4πε₀)e²/r = 1.44 MeV·fm / 3.0 fm = 0.48 MeV
In joules: 0.48 × 1.602 × 10−13 = 7.69 × 10−14 J
Temperature: (3/2)kBT = U, so T = 2U/(3kB) = (2 × 7.69 × 10−14)/(3 × 1.38 × 10−23) = 3.7 × 109 K
The Sun’s core is only about 1.5 × 107 K, roughly 250 times cooler. Fusion still proceeds there because of the high-energy tail of the speed distribution and quantum tunnelling through the barrier — but only very slowly, which is precisely why the Sun is still here.
Q11. Given 235U = 235.043930 u and 238U = 238.050788 u, which isotope has the greater total binding energy, and which is the more stable? Explain the apparent contradiction.
Uranium-235 (Z = 92, N = 143): parts = 92(1.007825) + 143(1.008665) = 236.958995 u; Δm = 1.915065 u; B.E. = 1783.88 MeV; B.E./A = 1783.88/235 = 7.591 MeV.
Uranium-238 (Z = 92, N = 146): parts = 92(1.007825) + 146(1.008665) = 239.984990 u; Δm = 1.934202 u; B.E. = 1801.71 MeV; B.E./A = 1801.71/238 = 7.570 MeV.
Answer: uranium-238 has the greater total binding energy (by 17.83 MeV), but uranium-235 has the greater binding energy per nucleon and is therefore the marginally more tightly bound arrangement.
There is no contradiction: U-238 simply has three more nucleons to bind. Total binding energy rewards size; binding energy per nucleon measures quality of binding. Stability comparisons always use the per-nucleon figure.
Q12. (Multi-part, exam style) Uranium-235 has B.E./A = 7.591 MeV and its fission fragments average 8.5 MeV per nucleon. (a) Estimate the energy released per fission. (b) How many fissions per second are needed for a 200 MW output at 100 per cent efficiency? (c) State, with a reason, whether the same argument would work for iron-56.
(a) All 235 nucleons climb by (8.5 − 7.591) = 0.909 MeV each, so
E = 235 × 0.909 = 213.6 MeV ≈ 214 MeV, comfortably consistent with the standard figure of about 200 MeV once you allow for the energy carried off by neutrons and for fragments whose B.E./A is a little under 8.5.
(b) Take 200 MeV per fission = 200 × 1.602 × 10−13 = 3.204 × 10−11 J.
Rate = P/E = (2.0 × 108 J s−1)/(3.204 × 10−11 J) = 6.24 × 1018 fissions per second.
(c) No. Iron-56 sits at 8.790 MeV per nucleon, at the very top of the curve, so any fragments would have a lower B.E./A. The climb would be negative, meaning you would have to supply energy — about 56 × (8.790 − 8.448) ≈ 19 MeV to split it into two silicon-28 nuclei. Iron will neither fission nor fuse profitably; it is the summit of the Iron Hill.
If you got at least nine of those twelve right on the first pass, you are in good shape for this chapter. If you got fewer, look at which ones went wrong — almost certainly they cluster in one place, either the mass-defect bookkeeping or the direction of energy flow, and one focused half hour will fix it.
And whenever a question in this chapter makes your mind go blank, remember where the answer lives. Draw the hill. Mark where the nucleons start and where they finish. If they moved up, energy came out. Both ends run to iron — and now you know why. More chapters, all written this way, are waiting in the study notes library.