Beta Decay & the Neutrino

Beta decay is one of the most useful processes physics knows. Radiocarbon dating rests entirely on the β⁻ decay of carbon-14 met above: living things maintain a fixed ratio of $^{14}$C to ordinary carbon, and once they die the $^{14}$C ticks down with a half-life of $5730$ years, so the residual beta activity reads off the age of wood, cloth, bone and charcoal up to about fifty thousand years old. Positron emission tomography (PET) turns β⁺ decay into a medical camera: a patient is given a sugar tagged with fluorine-18, and each emitted positron annihilates with a nearby electron into two $511\ \text{keV}$ gamma rays flying apart back-to-back, whose coincident arrival pinpoints where in the body the tracer collected — lighting up tumours and active brain tissue. Nuclear reactors are prolific beta emitters: their fission fragments are neutron-rich and shed beta particles for years, which is both the source of a reactor's dangerous "decay heat" after shutdown and the flood of antineutrinos that made the 1956 detection possible. And the neutrino itself, once an apologetic fudge, is now a tool: neutrino telescopes buried in Antarctic ice and deep mines watch the Sun's fusion core and distant supernovae through the one messenger that escapes them unimpeded. The particle Pauli feared could never be seen is now caught by the trillion.

Alpha decay throws a whole ready-made helium nucleus out of an over-heavy nucleus. Beta decay is subtler and, at bottom, stranger: nothing is thrown out that was inside to begin with. Instead one of the nucleons transforms into the other kind, and the electron (or positron) that flies off is created at the instant of decay — it was never an occupant of the nucleus, any more than the sound of a bell was hiding inside the metal before it was struck. There are no electrons living inside a nucleus; the beta electron is manufactured on the spot out of the energy released when a neutron becomes a proton. Why would a nucleus bother? Because it is trading a less stable arrangement for a more stable one. A nucleus with too many neutrons for its number of protons is like a see-saw loaded on one side; converting a surplus neutron into a proton (β⁻) evens the balance and drops the whole system to lower energy. A nucleus with too many protons does the reverse, converting a proton into a neutron (β⁺, or its cousin electron capture). In every case the mass number $A = Z + N$ stays the same — one nucleon simply changes its identity — while the atomic number $Z$ steps by one. On the chart of the nuclides the daughter sits one square across from the parent, never up or down the mass ladder.

The particle that emerges is a genuine electron (for β⁻) or its antimatter twin the positron (for β⁺): same mass $m_e$, opposite charge. But it never comes alone. Riding out beside it is the neutrino — neutral, of vanishingly small mass, so weakly interacting that a wall of lead a light-year thick would stop only half of a beam of them. That ghost is what makes the accounting work, and the rest of this article is the story of how a graph of electron energies proved it had to be there.

At the level of the individual nucleon there are two decay channels, plus a closely related third.

Beta-minus (β⁻) decay. A neutron converts to a proton. Charge must balance, so a negative electron is created; a fourth conservation law we meet below demands an antineutrino too:

For a whole nucleus $^{A}_{Z}X$ the newly minted proton stays behind and lifts the atomic number by one:

where: $^{A}_{Z}X$ is the parent nucleus, $^{A}_{Z+1}Y$ the daughter one place higher in the periodic table, $e^-$ the emitted beta electron and $\bar\nu_e$ the electron antineutrino. The physical reading: neutron-rich nuclei climb toward stability by turning a neutron into a proton. The canonical school example is carbon-14, $^{14}_{6}\text{C} \to\ ^{14}_{7}\text{N} + e^- + \bar\nu_e$, the reaction that dates ancient wood and bone.

Beta-plus (β⁺) decay. For a proton-rich nucleus the traffic runs the other way: a proton becomes a neutron, throwing out a positron (the electron's positive antiparticle) and an ordinary neutrino:

where: $e^+$ is the positron and $\nu_e$ the electron neutrino. Now $Z$ falls by one. A free proton cannot do this — it is lighter than a neutron, so the conversion would need energy from nowhere — but inside a proton-rich nucleus the gain in nuclear binding pays the bill. Positron emitters such as fluorine-18 are the heart of the PET scanner.

Electron capture. When a proton-rich nucleus does not have quite enough spare energy for β⁺ (we will see it needs $1.02\ \text{MeV}$), it can instead swallow one of its own inner atomic electrons: $p + e^- \to n + \nu_e$. The result — $Z \to Z-1$, $A$ fixed — is the same as β⁺, but only a single neutrino leaves the atom. It is beta decay's quiet third door and a favourite exam contrast. In every one of these, the total electric charge and the mass number $A$ are conserved, and exactly one lepton–neutrino pair is created. What is not conserved is the identity of the nucleon: beta decay is the one process a school student meets in which a particle literally turns into a different particle.

Beta decay, like every decay, is powered by a tiny loss of mass. The $Q$-value is the total energy released, $Q = \Delta m\,c^2$, shared as kinetic energy among the products (the recoiling daughter, the beta particle and the neutrino). A decay proceeds spontaneously only if $Q > 0$, i.e. only if the parent is heavier than the sum of its products. Here lurks the single most useful bookkeeping trick in nuclear physics: use atomic masses, not bare nuclear masses, and the electron masses cancel by themselves. A table of atomic masses already includes each atom's full complement of orbital electrons; because beta decay changes $Z$ by one, the electron count differs between parent and daughter atoms by exactly the one electron that is created, so it drops out of the sum. The two derivations below make this precise. The upshot for β⁻ is beautifully clean:

where: the $M$ are neutral-atom masses (in $\text{u}$, then multiplied by $931.5\ \text{MeV/u}$). No electron mass appears at all. For β⁺, by contrast, the same careful accounting leaves behind two electron masses:

where: $2m_e c^2 = 1.02\ \text{MeV}$ is the fixed penalty a positron emitter must overcome. This asymmetry between (4) and (5) is not a typo and not a curiosity — it is a standard mark-earner, and getting it right is what separates a correct β⁺ answer from a wrong one.

Now the puzzle that started it all. If beta decay were a two-body split — daughter plus electron only — then energy and momentum conservation would fix the electron's energy to a single sharp value, exactly as they do for the alpha particle. Fire off the same decay a million times and every electron should carry the identical energy $Q$ (minus the trifling recoil). That is emphatically not what James Chadwick found in 1914. The beta electrons came out with a continuous range of energies, from nearly zero all the way up to a definite maximum — the endpoint energy — with a broad hump in between. Most electrons carried far less than the full $Q$; the endpoint alone equalled it.

A continuous spectrum with a sharp ceiling can only mean one thing: the energy $Q$ is being shared among more than two products, and the share going to the electron varies from event to event. When the electron happens to take almost everything, its energy approaches the endpoint $Q$; when it takes almost nothing, the other product carries $Q$ away instead. That silent partner is the neutrino. It is always emitted; we simply cannot see it. The endpoint of the spectrum is therefore a direct measurement of $Q$, and its value is exactly what equation (4) predicts from the atomic masses. Energy was not the only law in danger. Spin was too. A neutron, a proton and an electron each have spin $\tfrac12$. A decay $n \to p + e^-$ would take a spin-$\tfrac12$ particle into two spin-$\tfrac12$ particles, whose spins can only add to $0$ or $1$ — never back to a half-integer. Angular momentum simply would not balance. Adding a third spin-$\tfrac12$ particle, the neutrino, fixes both crimes at once: it restores the half-integer total spin and it carries off the missing energy and momentum. Pauli's single new particle closed three conservation ledgers simultaneously, which is why, however desperate it seemed in 1930, it was almost certainly right. It took until 1956 for Cowan and Reines to catch antineutrinos streaming from a nuclear reactor and prove the ghost real.

Board level — carbon-14 decay and its Q-value (CBSE and ICSE pattern)

Carbon-14 is a beta-minus emitter. Write its decay equation, identify the daughter nuclide, and find the energy released, given the atomic masses $M(^{14}\text{C}) = 14.003242\ \text{u}$ and $M(^{14}\text{N}) = 14.003074\ \text{u}$, with $1\ \text{u} = 931.5\ \text{MeV}/c^2$.

Step 1. In β⁻ decay a neutron becomes a proton, so $Z$ rises by one ($6 \to 7$) while $A = 14$ is unchanged. Carbon ($Z=6$) becomes nitrogen ($Z=7$): $^{14}_{6}\text{C} \to\ ^{14}_{7}\text{N} + e^- + \bar\nu_e$. The daughter is $^{14}_{7}\text{N}$.

Step 2. For β⁻ the electron masses cancel, so use atomic masses directly (equation 4): $\Delta m = M(^{14}\text{C}) - M(^{14}\text{N}) = 14.003242 - 14.003074 = 0.000168\ \text{u}$.

Step 3. Convert to energy: $Q = \Delta m \times 931.5 = 0.000168 \times 931.5 = \boxed{0.156\ \text{MeV}}$.

Step 4. Interpret. This $0.156\ \text{MeV}$ is the maximum kinetic energy any emitted electron can have — the endpoint of the continuous spectrum. Most electrons carry less; on average only about a third of it, roughly $0.05\ \text{MeV}$, with the antineutrino taking the balance. The nitrogen nucleus, being far heavier, recoils with negligible energy.

Competitive level — fluorine-18, the PET tracer (JEE and NEET pattern)

Fluorine-18 decays by positron emission. Write the equation, and using $M(^{18}\text{F}) = 18.000938\ \text{u}$, $M(^{18}\text{O}) = 17.999160\ \text{u}$ and $m_e c^2 = 0.511\ \text{MeV}$, find the maximum kinetic energy of the emitted positron. Why must you subtract two electron masses, and what would happen if the mass difference were below $1.02\ \text{MeV}$?

Step 1. In β⁺ decay a proton becomes a neutron, so $Z$ falls by one ($9 \to 8$), $A = 18$ fixed. Fluorine ($Z=9$) becomes oxygen ($Z=8$): $^{18}_{9}\text{F} \to\ ^{18}_{8}\text{O} + e^+ + \nu_e$.

Step 2. Use the β⁺ formula (equation 5), $Q = [M(^{18}\text{F}) - M(^{18}\text{O}) - 2m_e]c^2$. The atomic-mass difference is $18.000938 - 17.999160 = 0.001778\ \text{u}$. The two-electron term is $2m_e = 2 \times 0.000549 = 0.001098\ \text{u}$.

Step 3. Subtract and convert: $Q = (0.001778 - 0.001098) \times 931.5 = 0.000680 \times 931.5 = \boxed{0.633\ \text{MeV}}$. This is the endpoint (maximum) positron energy; the neutrino shares the rest.

Step 4. The reasoning behind the $-2m_e$: the daughter oxygen atom is born with one electron too many (its nucleus lost a proton), and a positron of mass $m_e$ is created as well — two electron masses' worth of extra mass must be paid for out of $Q$. If the parent–daughter atomic-mass difference were less than $2m_e c^2 = 1.02\ \text{MeV}$, β⁺ would be forbidden and the nucleus would instead reach $^{18}\text{O}$ by capturing an inner atomic electron. (In fact $^{18}$F does a few percent of its decays by electron capture even so.) Timing note: about 90 seconds under exam pressure. The whole item is one subtraction and one multiplication — the only place to lose a mark is forgetting the $-2m_e$, which is exactly what the examiner is testing. Commit "β⁻ cancels, β⁺ costs two electron masses" to reflex and this class of question is free.

Competitive level — the free neutron and why the proton is stable (JEE Advanced pattern)

A free neutron is slightly heavier than a free proton: $m_n = 1.008665\ \text{u}$, $m_p = 1.007276\ \text{u}$, $m_e = 0.000549\ \text{u}$. Using particle (not atomic) masses, find the Q-value of free-neutron beta decay $n \to p + e^- + \bar\nu_e$, and explain why an isolated proton cannot beta-decay.

Step 1. Here there are no orbital electrons, so use the bare particle masses directly: $\Delta m = m_n - m_p - m_e = 1.008665 - 1.007276 - 0.000549 = 0.000840\ \text{u}$.

Step 2. Convert: $Q = 0.000840 \times 931.5 = \boxed{0.782\ \text{MeV}}$. Because $Q > 0$, a free neutron is unstable — and indeed it decays with a half-life of about $10$ minutes, the reason free neutrons are rare in nature.

Step 3. For the reverse, $p \to n + e^+ + \nu_e$, the mass change is $m_n - m_p + m_e > 0$: the products are heavier than the proton by $(m_n - m_p + m_e)c^2 \approx 1.81\ \text{MeV}$, so $Q < 0$ and the decay cannot happen. A free proton is stable; only inside a proton-rich nucleus, where nuclear binding energy supplies the deficit, can a bound proton convert to a neutron. Timing note: under a minute. The single insight worth memorising is the sign of $m_n - m_p$: the neutron is the heavier nucleon by about $1.29\ \text{MeV}/c^2$, which is why neutrons decay and protons (on their own) do not. Every free-neutron question hangs off that one fact.

Forgetting the $2m_e$ in β⁺. For β⁻ the electron masses cancel and $Q = [M(\text{parent}) - M(\text{daughter})]c^2$; for β⁺ you must subtract $2m_e c^2 = 1.02\ \text{MeV}$. Using the β⁻ formula for a positron emitter overstates $Q$ by exactly $1.02\ \text{MeV}$ and is the single most common β-decay error on JEE, NEET and A-Level. Anchor it: β⁻ cancels, β⁺ costs two.

Mixing up atomic and nuclear masses. Data tables almost always list atomic (neutral-atom) masses. The whole cancellation trick assumes you use them consistently. If a problem hands you bare nuclear masses, you must put the electron mass back in by hand. State which you are using.

Getting the direction of $Z$ wrong. β⁻ raises $Z$ by one (neutron $\to$ proton); β⁺ and electron capture lower it. The mass number $A$ never changes in beta decay. Writing the daughter one place too low for β⁻ is an easy, avoidable slip.

Reading the spectrum as a single energy. The beta particle does not come out at a fixed energy. Only the endpoint of its continuous spectrum equals $Q$; the average is far lower (about a third of $Q$). A question that asks for "the energy of the beta particle" is usually asking for the maximum, i.e. $Q$ — say so explicitly.

Calling the neutrino massless-and-chargeless "nothing." The neutrino is a real particle: neutral, spin $\tfrac12$, with a tiny but non-zero mass. It is needed to conserve energy, momentum and angular momentum. Do not omit it from a decay equation, and do not confuse $\nu_e$ (emitted in β⁺) with $\bar\nu_e$ (emitted in β⁻).

Board scope. The identity of β⁻/β⁺ decay and the $Z, A$ changes are core on CBSE, ICSE, JEE, NEET, AP Physics 2, IB HL, A-Level 9702 and even IGCSE 0625 (qualitatively). The $Q$-value in MeV, the $931.5\ \text{MeV/u}$ conversion and the $2m_e$ subtlety are not on IGCSE — that board treats beta decay only descriptively. The continuous spectrum and the neutrino argument sit at the higher tier (JEE Advanced, A-Level, IB HL); NEET and CBSE expect you to know the neutrino exists and why, but not to compute its spectrum.

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the radioactivity and beta-decay sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Krane, Introductory Nuclear Physics, Chapter 9, for the historical and theoretical route to the neutrino · Serway/Halliday modern-physics chapters for the $Q$-value conventions.

Provenance. Standard nuclear data; atomic masses from the AME table. Recompute: β⁻ Q-value formula $Q = [M(\text{parent}) - M(\text{daughter})]\,931.5\ \text{MeV}$; β⁺ formula with the $-2m_e$ term; $^{14}\text{C} \to\ ^{14}\text{N}$ mass difference $0.000168\ \text{u}$ giving $Q = 0.156\ \text{MeV}$; average β energy $\approx Q/3 \approx 0.05\ \text{MeV}$; $^{18}\text{F} \to\ ^{18}\text{O}$ mass difference $0.001778\ \text{u}$, minus $2m_e = 0.001098\ \text{u}$, giving $Q = 0.633\ \text{MeV}$; $m_e c^2 = 0.511\ \text{MeV}$ and $2m_e c^2 = 1.02\ \text{MeV}$; free-neutron $\Delta m = 0.000840\ \text{u}$ giving $Q = 0.782\ \text{MeV}$; neutron–proton mass difference $(m_n - m_p)c^2 = 1.29\ \text{MeV}$; forbidden free-proton decay margin $\approx 1.81\ \text{MeV}$.

Prerequisites. Radioactivity: Alpha, Beta & Gamma — the three radiations compared, of which beta is the middle one this article opens up. Its ideas of nuclear transmutation and the $Q$-value carry straight over. Leads to. Gamma Decay & Nuclear Energy Levels, since a beta decay often leaves the daughter nucleus excited, and it sheds that surplus as a gamma photon; and The Radioactive Decay Law, which gives the rate at which these decays happen. For enrichment, the university seam Particle Decay Kinematics in Special Relativity (go-deeper only, never an exam dependency here) treats the energy-momentum sharing that makes the beta spectrum continuous.