Balancing nuclear equations

Every nucleus is described by two numbers. Once you can read them, every decay in this lesson is the same bookkeeping exercise.

The mass number $A$ is protons + neutrons, written top-left. It is conserved — nucleons do not appear or vanish here, they just regroup. The atomic number $Z$ is protons, written bottom-left. It is conserved as charge, and it decides which element you have. Change $Z$ and you change the element itself.

${}^{A}_{Z}X$

Take uranium-238, ${}^{238}_{92}\mathrm{U}$ . It has 92 protons, and $238 - 92 = 146$ neutrons. If a decay removes 2 protons, $Z$ falls to 90 and the atom is no longer uranium at all — it is thorium. That is the whole game: track $A$ down each column, track $Z$ down each column, and read off the new element.

Worked problem 1 — alpha decay

A heavy nucleus throws out a chunk of 2 protons + 2 neutrons — a helium nucleus, ${}^{4}_{2}\mathrm{He}$ . Because the chunk carries away 4 nucleons, $A$ drops by 4, and because it carries 2 protons, $Z$ drops by 2 and the element slides two places down the periodic table.

Uranium-238 decays by alpha emission. Write the equation, then check both numbers.

${}^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}$

Checking $A$ : $238$ on the left, $234 + 4 = 238$ on the right. Checking $Z$ : $92$ on the left, $90 + 2 = 92$ on the right. Both balance, so the daughter must be thorium-234. About $4.27\ \mathrm{MeV}$ of missing mass becomes the kinetic energy that launches the alpha at ~5% of light speed.

Try it. Radium-226 ( ${}^{226}_{88}\mathrm{Ra}$ ) emits an alpha. Subtract 4 from $A$ and 2 from $Z$ : you should get ${}^{222}_{86}\mathrm{Rn}$ , radon-222. If your $Z$ gives a different element than your textbook says, recheck the subtraction before doubting the table.

Worked problem 2 — beta-minus decay

Here no chunk leaves. A neutron becomes a proton, so $A$ does not change at all and $Z$ goes up by 1. At quark level a down quark turns into an up quark, and the released energy is manufactured into an electron and an antineutrino that never existed before:

$d \to u + e^- + \bar{\nu} \qquad n \to p + e^- + \bar{\nu}$

The classic case is carbon dating. Carbon-14 decays to nitrogen-14:

${}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + e^- + \bar{\nu} + 0.156\ \mathrm{MeV}$

Read it as sentences. Mass number: $14$ on the left, $14$ on the right — nothing was lost. Charge: $6$ on the left, $7 - 1 = 6$ on the right once you count the electron’s $-1$ . The element moved up one place, carbon to nitrogen, exactly as $Z + 1$ predicts. The $0.156\ \mathrm{MeV}$ is shared at random between the electron and antineutrino, which is why beta electrons come out with a spread of energies.

Try it. Phosphorus-32 ( ${}^{32}_{15}\mathrm{P}$ ) is a beta-minus emitter used in biology labs. Keep $A = 32$ , raise $Z$ to 16: you get ${}^{32}_{16}\mathrm{S}$ , sulfur-32, plus $e^- + \bar{\nu}$ . Any answer that changes $A$ is wrong for beta decay — that is the signature to check first.

Worked problem 3 — beta-plus decay

This is the mirror image. A proton becomes a neutron, so again $A$ is unchanged and $Z$ falls by 1. Out comes a positron — the antimatter twin of an electron — plus a neutrino:

$p \to n + e^+ + \nu$

Fluorine-18, the PET-scan isotope with a 110-minute half-life, decays this way:

${}^{18}_{9}\mathrm{F} \to {}^{18}_{8}\mathrm{O} + e^+ + \nu$

In words: mass stays 18, charge goes $9 \to 8 + 1$ , balanced by the positron’s $+1$ . Oxygen-18 is stable, so the chain stops. The positron itself does not travel far — within about a millimeter it meets an ordinary electron and both vanish into two gamma rays flying opposite directions:

$e^+ + e^- \to \gamma + \gamma \quad (511\ \mathrm{keV\ each})$

Those paired gammas are what a PET scanner detects. One catch governs the whole mode: manufacturing a positron costs at least $1.022\ \mathrm{MeV}$ , twice an electron’s mass. A lone proton can never beta-plus decay — only a nucleus with that much spare energy can afford it.

Try it. Sodium-22 ( ${}^{22}_{11}\mathrm{Na}$ ) emits a positron. Hold $A = 22$ , drop $Z$ to 10: neon-22, ${}^{22}_{10}\mathrm{Ne}$ . If the parent had less than $1.022\ \mathrm{MeV}$ to spare, this route would be closed and it would use electron capture instead.

Worked problem 4 — electron capture

Same net result as beta-plus ( $Z$ down by 1, $A$ unchanged) by a cheaper route. The nucleus absorbs one of its own inner electrons:

$p + e^- \to n + \nu$

Iron-55 is the textbook case:

${}^{55}_{26}\mathrm{Fe} + e^- \to {}^{55}_{25}\mathrm{Mn} + \nu$

Narrate the bookkeeping: $A$ stays 55. Charge goes $26 - 1 = 25$ on the left (counting the captured electron), matching manganese’s 25. Only a neutrino leaves, so there is no energy-sharing spread and no $1.022\ \mathrm{MeV}$ threshold — any spare energy above zero will do. The visible evidence is indirect: the innermost electron shell now has a hole, an outer electron drops in, and the difference comes out as X-rays ( $5.9\ \mathrm{keV}$ for manganese).

Try it. Potassium-40 ( ${}^{40}_{19}\mathrm{K}$ ) captures an electron about 11% of the time. The answer is ${}^{40}_{18}\mathrm{Ar}$ — argon-40, which is where nearly all the argon in Earth’s air came from. Note the pattern: beta-plus and electron capture give the same daughter for the same parent. The difference is only what flies out and how much energy the parent needed.

How to check any answer

Run this checklist before trusting a balanced equation. First, does $A$ match? Alpha changes it by 4; every other mode here leaves it identical. Second, does $Z$ match once emitted particles are counted ( $e^-$ as $-1$ , $e^+$ as $+1$ )? Third, does the $Z$ shift match the mode ( $-2$ alpha, $+1$ beta-minus, $-1$ beta-plus/EC)? If all three agree, look up $Z$ in the periodic table — that element name is part of the answer.

Decay	What changes	$A$	$Z$	Comes out
Alpha	loses 2p + 2n	$-4$	$-2$	${}^{4}_{2}\mathrm{He}$
Beta-minus	$n \to p$	same	$+1$	$e^- + \bar{\nu}$
Beta-plus	$p \to n$	same	$-1$	$e^+ + \nu$
Electron capture	$p + e^- \to n$	same	$-1$	$\nu$ only

Next: alpha decay, step by step — we dive straight in with the mechanism, no quark changes.