Inside the Nucleus: Protons, Neutrons & Isotopes

The two-ingredient nucleus is the foundation under an entire industry of measurement and technology. Deuterium — hydrogen's one-neutron isotope — labels molecules so their fate can be traced, and heavy water ($\text{D}_2\text{O}$) moderates neutrons in reactors. The uneven neutron counts of isotopes are read directly by mass spectrometry, which sorts atoms by mass to fingerprint everything from pollutants to ancient climates locked in ice cores. Chadwick's neutron itself became a workhorse: slow neutron beams image the hydrogen in living tissue and engine parts where X-rays fail, and neutron activation reveals trace elements in a painting or a moon rock. Every nuclear reactor runs on the fact that a neutron, being uncharged, walks straight into a nucleus that would repel any proton — the same neutrality that made it so hard to discover is exactly what makes it useful. And the ratio of neutrons to protons that keeps a nucleus stable, first glimpsed here, is the organising principle of the whole chart of the nuclides that the rest of this chapter explores.

Start from where the previous article left off: Rutherford's atom is a swarm of light electrons around a minuscule, heavy, positive nucleus. The electrons are settled — they carry the chemistry and the spectra. The mystery is the nucleus itself, and by the late 1920s the evidence had narrowed it to a paradox. The nuclear charge is exactly $+Ze$, where $Z$ is the element's place in the periodic table, so the nucleus must contain $Z$ protons, each of charge $+e$ and mass about $1\,\text{u}$. But the nuclear mass is close to $A \approx 2Z$ atomic mass units for light elements and even more lopsided for heavy ones — uranium has $Z=92$ but weighs $238\,\text{u}$. The protons account for the charge and only about half the mass. The missing half is neutral (or the charge would be wrong) and heavy (about a proton's mass each). For a while physicists patched the gap with a wrong guess: perhaps the nucleus held $A$ protons to make the mass, plus $A - Z$ electrons tucked inside to cancel the surplus charge back down to $+Ze$. This "proton–electron" model was popular but fatally broken — it clashed with the newly-minted quantum mechanics (an electron confined to a nucleus a few femtometres across would, by the uncertainty principle, need far more energy than any nucleus contains) and it predicted the wrong nuclear spins and magnetic moments. The honest answer was simpler and stranger: a second, genuinely new particle — neutral, with essentially the proton's mass. Rutherford had guessed as much in his 1920 Bakerian Lecture and even named it in advance, the neutron. It took twelve more years to prove it existed.

The picture to carry forward is deliberately plain: a nucleus is a tight ball of protons and neutrons — nucleons — with nothing else inside. The protons are what make it an atom of a particular element and what give it its charge; the neutrons are dead weight electrically, contributing mass and, as we will see, the glue budget that lets the pack survive its own repulsion.

The neutron is hard to find for exactly the reason it is useful: it is electrically neutral, so it ionises almost nothing, curves in no magnetic field, and sails through matter leaving no track. The trail that led to it began in 1930, when Walther Bothe and Herbert Becker bombarded beryllium with alpha particles from polonium and found a new, intensely penetrating radiation coming off — penetrating enough to pass through lead. They assumed it was high-energy gamma radiation, since gamma rays were the only known neutral, penetrating radiation. In 1932 Irène Joliot-Curie and Frédéric Joliot passed this "beryllium radiation" into paraffin wax (which is rich in hydrogen) and watched it eject fast protons — and they too blamed energetic gamma rays. Chadwick saw the flaw immediately. For a gamma photon to knock a proton forward at the observed speed by a Compton-style collision, the photon would need an energy of around $50\,\text{MeV}$ — vastly more than the beryllium reaction could possibly supply. Worse, the same radiation also knocked much heavier nitrogen nuclei forward, and no single gamma energy could account for both recoils at once. The gamma hypothesis violated energy and momentum conservation. Chadwick proposed instead that the beryllium radiation was a stream of neutral particles of about the proton's mass — Rutherford's long-predicted neutron — produced by the reaction $^{9}_{4}\text{Be} + \, ^{4}_{2}\text{He} \rightarrow \, ^{12}_{\ 6}\text{C} + \, ^{1}_{0}\text{n}$. A heavy neutral particle transfers momentum in a plain mechanical collision, no impossible energies required.

How do you weigh a particle you cannot see? By the collision it makes. When a moving object of mass $m$ strikes a stationary object of mass $M$ head-on in an elastic collision, the struck object is driven forward with a top speed set purely by the mass ratio:

where: $V$ is the incoming particle's speed, $m$ its (unknown) mass, $M$ the target's mass, and $u_M$ the maximum speed the target can pick up. The physical reading is the same intuition a pool player has: a heavy ball drives a light one to nearly twice its own speed, a light ball barely nudges a heavy one, and equal masses trade velocity completely — if $m = M$, equation (1) gives $u_M = V$. The neutron's mass is locked inside that ratio, and comparing how hard it kicks two different targets — light protons and heavier nitrogen nuclei — lets us solve for it without ever knowing the neutron's own speed.

With two building blocks named, the nucleus becomes a matter of counting. Three integers describe any nuclear species — any nuclide — completely:

where: $Z$ is the atomic number, $N$ the neutron number, and $A$ the mass number — the total count of nucleons. The physical reading of equation (3) is the load-bearing fact of the whole chapter: $Z$ alone decides which element you have, because it fixes the nuclear charge $+Ze$ and therefore how many electrons the neutral atom carries and thus all of its chemistry; $N$ is free to vary without changing the element; and $A$, being just the nucleon count, is very nearly the atomic mass in atomic mass units, because each nucleon weighs about $1\,\text{u}$. The compact way to record all this is the nuclide symbol $^{A}_{Z}\text{X}$, where $\text{X}$ is the chemical symbol, the mass number $A$ rides as a superscript and the atomic number $Z$ as a subscript on the left. Uranium's common isotope is $^{238}_{\ 92}\text{U}$: ninety-two protons ($Z=92$), and $N = A - Z = 238 - 92 = 146$ neutrons. Because the symbol $\text{U}$ already means $Z = 92$, the subscript is redundant and often dropped — you will see $^{238}\text{U}$ or "U-238" and are expected to recover $Z$ from the periodic table. A gold nucleus $^{197}_{\ 79}\text{Au}$ has $79$ protons and $197 - 79 = 118$ neutrons. The special case is hydrogen, $^{1}_{1}\text{H}$: a single proton, no neutrons at all — the only nucleus that is a bare proton. One more number is worth fixing in your head. The atomic mass unit is defined by carbon: one atom of $^{12}\text{C}$ weighs exactly $12\,\text{u}$, so $1\,\text{u}$ is one-twelfth of a carbon-12 atom, equal to $1.66\times10^{-27}\,\text{kg}$. On this scale the proton is $1.007276\,\text{u}$, the neutron $1.008665\,\text{u}$, and the electron a mere $0.000549\,\text{u}$ — about $1/1836$ of a proton. That last figure is why the nucleus holds "almost all" of the atom's mass: in a hydrogen atom the electron is only $0.000549/1.007825 \approx 0.05\%$ of the total, so the nucleus carries more than $99.9\%$ of the weight while occupying less than a trillionth of the volume.

Because $N$ can change while $Z$ stays fixed, a single element comes in several nuclear versions of different mass. These are its isotopes — same $Z$, same chemistry, same place in the periodic table, but different $N$ and hence different $A$. Hydrogen is the textbook trio: ordinary hydrogen $^{1}_{1}\text{H}$ (one proton, no neutrons), deuterium $^{2}_{1}\text{H}$ (one proton, one neutron), and tritium $^{3}_{1}\text{H}$ (one proton, two neutrons). All three are hydrogen — one proton, one electron, identical chemistry — but they weigh $1$, $2$ and $3$ units. Carbon likewise appears as stable $^{12}\text{C}$ and $^{13}\text{C}$ and radioactive $^{14}\text{C}$, with $6$, $7$ and $8$ neutrons respectively; the last is the clock behind radiocarbon dating, a later article.

Isotopes are why the atomic masses on the periodic table are not whole numbers. Chlorine is quoted as about $35.5\,\text{u}$ — not because any chlorine atom weighs that, but because natural chlorine is roughly $75.77\%$ $^{35}\text{Cl}$ and $24.23\%$ $^{37}\text{Cl}$, and the weighted average $0.7577\times35 + 0.2423\times37 = 35.5$ lands between the two. The periodic-table mass is a population average over a nuclide mixture. Two companion terms round out the vocabulary, distinguished by which number is shared. Isobars are nuclei with the same mass number $A$ but different $Z$ — different elements of equal weight, such as $^{40}_{18}\text{Ar}$, $^{40}_{19}\text{K}$ and $^{40}_{20}\text{Ca}$, all with $A = 40$. Isotones share the neutron number $N$ but differ in $Z$ — for instance $^{13}_{\ 6}\text{C}$ and $^{14}_{\ 7}\text{N}$ both have $N = 7$. A tidy way to remember the trio: isotopes match on protons (same element — the "p" hides in the definition, same $Z$), isobars match on the big number $A$, and isotones match on neutrons $N$.

There is a serious problem with the picture we have just built, and every good student should feel it: a nucleus is a cluster of positive protons packed into a space a few femtometres across, and like charges repel. Squeeze two protons that close and the electric repulsion is not a nudge — it is enormous. The Coulomb potential energy of two protons a distance $r$ apart is

where: $k = 8.99\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ is the Coulomb constant, $e$ the elementary charge, and the combination $ke^2$ works out to the handy nuclear shortcut $1.44\ \text{MeV}\,\cdot\,\text{fm}$. The physical reading: put two protons a typical nuclear separation of about $2\ \text{fm}$ apart and equation (4) gives $U = 1.44/2 \approx 0.72\ \text{MeV}$ of repulsive energy per pair — and this is where nuclear physics parts company with chemistry. Chemical bond energies are a few electronvolts; here we are dealing with millions of electronvolts, a factor of about $10^{6}$ larger. If electricity were the only force in play, no nucleus beyond a single proton could exist for an instant; the protons would fly apart explosively. Since nuclei plainly do exist, there must be another force — attractive, acting between nucleons regardless of charge, and strong enough to beat the Coulomb repulsion at femtometre range. This is the strong nuclear force. It has three defining features you should carry: it is short-ranged, effectively vanishing beyond about $2$–$3\ \text{fm}$, so it binds only immediate neighbours and does not reach across a large nucleus; it is charge-independent, pulling proton–proton, proton–neutron and neutron–neutron pairs alike; and at nuclear separations it is roughly a hundred times stronger than the electric repulsion, which is why it wins. This immediately explains why nuclei need neutrons at all — and why heavy nuclei need more neutrons than protons. Every proton you add contributes both strong attraction (good, short-range) and Coulomb repulsion (bad, long-range, felt across the whole nucleus). Every neutron you add contributes strong attraction with no extra repulsion — pure glue. Light nuclei can get by with $N \approx Z$ (helium is $2$ and $2$, carbon $6$ and $6$), but as $Z$ climbs, the long-range Coulomb repulsion between all the protons grows faster than the short-range binding, and stability demands a growing surplus of neutrons to supply extra glue without extra charge. By uranium the balance has shifted to $92$ protons and $146$ neutrons — a neutron surplus of $54$. Push $Z$ high enough and even a neutron surplus cannot save it: the nucleus becomes unstable and decays, which is where the rest of this chapter goes.

Board level — reading a nuclide and finding an isotope (CBSE and ICSE pattern)

(a) For the nucleus $^{197}_{\ 79}\text{Au}$, state the number of protons, neutrons and nucleons. (b) An isotope of gold has two more neutrons. Write its symbol and its mass number. (c) A nucleus $^{40}_{19}\text{K}$ and a nucleus $^{40}_{20}\text{Ca}$ — what is their relationship?

Step 1. Gold has $Z = 79$, so $79$ protons. The mass number is $A = 197$, so the neutron count is $N = A - Z = 197 - 79 = \boxed{118}$ neutrons, and the nucleon count is $A = \boxed{197}$.

Step 2. "Two more neutrons" changes $N$ but not $Z$ — it is still gold, still $Z = 79$. The new mass number is $A = 197 + 2 = 199$, so the isotope is $\boxed{^{199}_{\ 79}\text{Au}}$. Both are gold because the proton number is unchanged.

Step 3. Potassium-40 and calcium-40 share the same mass number $A = 40$ but have different atomic numbers ($Z = 19$ versus $Z = 20$), so they are isobars — different elements of equal mass number.

Competitive level — the neutron mass from a recoil (JEE pattern; NEET uses only the elastic-collision special case)

A neutral particle of mass $m$ (in units of the proton mass) collides head-on and elastically with a stationary proton ($M_H = 1$), giving it a maximum recoil speed $u_H$; the same particle striking a stationary nitrogen nucleus ($M_N = 14$) gives it $u_N$. Measurements give $u_H / u_N = 7.0$. Find $m$, and state what particle it is. Then check the equal-mass limit.

Step 1. For an elastic head-on hit the target's maximum speed is $u = 2mV/(m + M)$ (equation 1). Writing it for each target: $u_H = 2mV/(m+1)$ and $u_N = 2mV/(m+14)$.

Step 2. Divide to cancel the unknown neutron speed $V$: $u_H/u_N = (m+14)/(m+1) = 7.0$.

Step 3. Solve the linear equation: $7.0(m+1) = m + 14 \Rightarrow 7.0m + 7.0 = m + 14 \Rightarrow 6.0m = 7.0 \Rightarrow m = \boxed{1.17}$ proton masses. Within experimental error this is one proton mass — the particle is the neutron.

Step 4. Equal-mass check: if $m = M$, then $u = 2mV/(2m) = V$ — the projectile stops dead and the target leaves with the projectile's full speed, the classic "Newton's cradle" transfer. This is why hydrogen (protons) makes the best neutron detector and the best moderator: a neutron hitting a proton can hand over all its energy in one bounce. Timing note: about 90 seconds. The whole item is one algebraic ratio — the moment you divide the two collision equations the two unknowns you cannot measure (the neutron's mass appears, but its speed $V$ and the common factor $2m$) cancel, and you are left with a single linear equation in $m$. Do not try to find $V$; it is designed to disappear. Recognising that equal masses give total velocity transfer earns the last mark for free.

Confusing $A$ with atomic mass. The mass number $A$ is an integer count of nucleons; the atomic mass is a measured weight in $\text{u}$, close to $A$ but not identical. Reserve "$238$" for the count and "$238.05\,\text{u}$" for the weight. Loses marks on CBSE, ICSE and A-Level definition questions.

Letting $Z$ change when only $N$ should. Adding or removing neutrons makes an isotope of the same element; the proton number $Z$ must stay put. Only a change in $Z$ changes the element. Watch this in every "write the isotope" or "balance the reaction" question — conserve both $A$ and $Z$.

Reading the periodic-table mass as a single nucleus. Chlorine's $35.5\,\text{u}$ is a population average of $^{35}\text{Cl}$ and $^{37}\text{Cl}$; no chlorine nucleus weighs $35.5$. Any question about a single nucleus uses whole-number $A$, not the tabulated average.

Thinking electrons are inside the nucleus. In $\beta^{-}$ decay an electron is emitted from the nucleus, which tempts students to picture electrons sitting there — but the modern nucleus contains only protons and neutrons; the emitted electron is created at the moment of decay (a later article). The old proton–electron nuclear model is wrong.

Forgetting why heavy nuclei are neutron-rich. It is not arbitrary: neutrons add strong-force glue without adding Coulomb repulsion, so as $Z$ grows the nucleus needs a neutron surplus ($N > Z$) to stay bound. A favourite "explain why" question on JEE and A-Level.

Board scope. The composition of the nucleus, $Z$/$N$/$A$, nuclide notation and isotopes are core on every board here — CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB (SL and HL) and A-Level 9702 — and even IGCSE 0625 carries the qualitative half (protons, neutrons, isotopes, the symbol). What IGCSE does not carry is the quantitative mass–energy machinery of the later articles (the $931.5\ \text{MeV}$ conversion, binding-energy sums, Q-values). The neutron-mass-from-recoil calculation, the strong-force estimate and the $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$ Coulomb calculation here are all enrichment — expected in full at JEE Advanced, IB HL and A-Level; IGCSE and NEET take the neutron's discovery and the strong-force story only qualitatively.

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the composition and size sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Halliday, Resnick and Walker, "Nuclear Physics" · Krane, Introductory Nuclear Physics, for the neutron-discovery history in detail.

Provenance. Standard nuclear data. Recompute: $^{238}\text{U}$ has $N = 238 - 92 = 146$; $^{197}\text{Au}$ has $N = 197 - 79 = 118$; proton $1.007276\,\text{u} = 938.3\ \text{MeV}$, neutron $1.008665\,\text{u} = 939.6\ \text{MeV}$, electron $0.000549\,\text{u} = 0.511\ \text{MeV}$, all via $\times 931.5$; neutron$-$proton gap $= 1.3\ \text{MeV}$; $m_p/m_e = 938.3/0.511 \approx 1836$; electron is $0.000549/1.007825 \approx 0.05\%$ of a hydrogen atom; $1\,\text{u} = 1.66\times10^{-27}\,\text{kg}$, $^{12}\text{C} \equiv 12\,\text{u}$; Chadwick recoil ratio $u_H/u_N = 3.3\times10^{7}/4.7\times10^{6} \approx 7.0 \Rightarrow m = (14-7.0)/(7.0-1) \approx 1.17$ proton masses, and equal-mass limit $u = V$; chlorine average $0.7577\times35 + 0.2423\times37 \approx 35.5$; $ke^2 = 1.44\ \text{MeV}\,\cdot\,\text{fm}$, $U = 1.44/2 = 0.72\ \text{MeV}$ at $2\ \text{fm}$, ratio to a $5\ \text{eV}$ bond $\approx 1.4\times10^{5}$; free-neutron half-life $\approx 10\ \text{min}$.

Prerequisites. Rutherford's Nuclear Atom — the scattering result that put the nucleus on the map and set up the question this article answers. Leads to. Nuclear Size and Density, which measures how small and how staggeringly dense this two-ingredient nucleus actually is, and from there the mass–energy and binding-energy articles that explain what holds it together quantitatively. For enrichment, the university nuclear-physics seam treats the strong force and the chart of nuclides in full (go-deeper only, never an exam dependency here).