Writing and Balancing Nuclear Equations

The two conservation rules for nuclear equations and worked examples of alpha, beta, positron, and gamma equations.

Quick answer

A nuclear equation shows a nuclear change with an arrow, and it must satisfy two conservation laws: the sum of the mass numbers A must be equal on both sides, and the sum of the atomic numbers Z must be equal on both sides. Individual elements are not conserved, because transmutation changes one element into another.

Example: Alpha decay of uranium-238: ^238_92 U -> ^234_90 Th + ^4_2 He. Mass: 238 = 234 + 4. Charge: 92 = 90 + 2. Both sides balance.

What you will learn

  • State the two conservation laws that every nuclear equation must obey
  • Write the balanced equation for an alpha decay
  • Write the balanced equation for a beta-minus decay
  • Identify positron emission and electron capture in equation form
  • Find a missing particle or product in an incomplete nuclear equation

The two conservation rules

A nuclear equation uses an arrow to show a nuclear transformation, for example:

92238U→90234Th+24He^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He

The equation is correct only if two quantities are conserved — equal on the left and right sides:

  1. Conservation of mass number: the sum of all A values on the left equals the sum of all A values on the right.
  2. Conservation of atomic number (charge): the sum of all Z values on the left equals the sum of all Z values on the right.

These two rules replace the atom-counting you use in chemistry. Because nuclear reactions change protons into neutrons and vice versa, the elements themselves are not conserved — and that is expected, not an error.

Writing an alpha decay equation

In alpha decay the parent ejects an alpha particle, 24He^{4}_{2}He. Therefore the daughter always has a mass number 4 lower and an atomic number 2 lower than the parent.

92238U→90234Th+24He^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He

Check: AA: 238=234+4238 = 234 + 4 ✓. ZZ: 92=90+292 = 90 + 2 ✓.

The daughter's identity follows from its Z: Z = 90 is thorium, so the product must be 90234Th^{234}_{90}Th. A frequent textbook example is radium-226 decaying to radon-222: 88226Ra→86222Rn+24He^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^{4}_{2}He.

Alpha shortcut

Alpha decay: subtract 4 from A, subtract 2 from Z, and identify the new element from the new Z. The emitted particle is always ^4_2 He.

Writing a beta-minus decay equation

In beta-minus decay, a neutron converts into a proton and ejects an electron, −10e^{0}_{-1}e. The mass number does not change (a neutron becomes a proton, so the nucleon count is the same), but the atomic number increases by one.

614C→714N+−10e^{14}_{6}C \rightarrow ^{14}_{7}N + ^{0}_{-1}e

Check: AA: 14=14+014 = 14 + 0 ✓. ZZ: 6=7+(−1)6 = 7 + (-1) ✓.

Because Z increases by one, the daughter is the next element up the periodic table: carbon-14 becomes nitrogen-14. Note how the negative Z of the electron makes the charge sum work — this is the detail students most often forget.

Positron emission, electron capture, and gamma

Two related processes complete the picture. In positron emission (beta-plus decay), a proton converts into a neutron and ejects a positron, +10e^{0}_{+1}e. The mass number is unchanged and Z decreases by one:

1122Na→1022Ne++10e^{22}_{11}Na \rightarrow ^{22}_{10}Ne + ^{0}_{+1}e

In electron capture, the nucleus absorbs one of its own inner electrons. A proton plus the captured electron becomes a neutron, so Z decreases by one with no emitted particle:

47Be+−10e→37Li^{7}_{4}Be + ^{0}_{-1}e \rightarrow ^{7}_{3}Li

Note that electron capture is written with the electron on the left side of the arrow — it is a reactant, not a product.

Finally, gamma emission is written with a 00γ^{0}_{0}\gamma term that contributes 0 to both sums. For example, an excited technetium-99 nucleus relaxes: 4399mTc→4399Tc+00γ^{99m}_{43}Tc \rightarrow ^{99}_{43}Tc + ^{0}_{0}\gamma. The 'm' marks the excited (metastable) state; A and Z are unchanged.

Finding a missing particle

When a piece of the equation is missing, let the conservation laws find it. Suppose radon-222 decays to polonium-218 plus an unknown particle:

86222Rn→84218Po+???^{222}_{86}Rn \rightarrow ^{218}_{84}Po + ^{?}_{?}?

  1. Mass number: 222−218=4222 - 218 = 4, so the unknown has A=4A = 4.
  2. Atomic number: 86−84=286 - 84 = 2, so the unknown has Z=2Z = 2.
  3. Identify: A=4A = 4, Z=2Z = 2 is 24He^{4}_{2}He — an alpha particle.

The completed equation is 86222Rn→84218Po+24He^{222}_{86}Rn \rightarrow ^{218}_{84}Po + ^{4}_{2}He. This difference method works for any missing particle: subtract the known products from the parent on both the A and Z lines.

A checklist for balanced equations

Work through these checks before calling any nuclear equation finished:

  1. Sum A on each side separately. Every reactant and product counts, including small particles like 01n^{1}_{0}n and −10e^{0}_{-1}e.
  2. Sum Z on each side separately. Remember that beta particles carry Z=−1Z = -1 and positrons Z=+1Z = +1.
  3. Verify each element matches its Z. If you wrote 90234Th^{234}_{90}Th, confirm that element 90 is actually thorium.
  4. Check the physical sense. Alpha decay lowers A by 4 and Z by 2; beta-minus raises Z by 1 with A unchanged. If your equation does something else, re-examine the decay mode.

With these four checks, balancing nuclear equations becomes a routine arithmetic exercise rather than a guessing game.

Variables and units

SymbolNameUnitMeaning
24He^4_2 Healpha particle—Particle with mass number 4 and atomic number 2, identical to a helium-4 nucleus.
−01e^0_-1 ebeta-minus particle—High-speed electron emitted in beta-minus decay; mass number 0, atomic number -1.
+01e^0_+1 epositron—Positron emitted in beta-plus decay; mass number 0, atomic number +1.
00γ^0_0 γgamma ray—High-energy photon; mass number 0 and atomic number 0, so it never affects the A or Z balance.

Common mistakes

✗ Balancing a nuclear equation means balancing atoms of each element like in chemistry.

Elements change identity in nuclear reactions, so you cannot balance element by element. Balance the total mass number A and the total atomic number Z separately; the elements then take care of themselves.

✗ A correct-looking equation with balanced atoms is automatically balanced.

Consider ^14_6 C -> ^14_7 N + ^0_-1 e: the 'atoms' are not balanced (one carbon becomes one nitrogen plus an electron), yet A (14 = 14 + 0) and Z (6 = 7 - 1) both balance. That is the correct criterion.

✗ Gamma emission changes the equation because energy leaves.

Gamma emission is written as ^0_0 gamma (or just gamma), with mass number 0 and charge 0. It contributes nothing to the A or Z sums, so the parent and daughter have identical A and Z.

Real-world applications

  • Predicting the products of an unknown radioactive decay

  • Identifying decay modes from parent-daughter data in decay series

  • Verifying proposed nuclear reactions in research and forensics

  • Determining what radiation type a source emits by completing its equation

Frequently asked questions

How is balancing a nuclear equation different from balancing a chemical equation?

Chemical equations balance atoms of each element; nuclear equations balance mass number A and atomic number Z separately. Atoms of individual elements are NOT conserved in nuclear equations, because one element can change into another. Also, nuclear equations track individual nuclei, not moles of compounds.

Are ^1_0 n and a written-out neutron the same thing?

Yes — both are written as ^1_0 n, a particle with mass number 1 and charge 0. They are the same particle; textbooks use whichever form fits the notation of the surrounding equation. The same applies to protons (^1_1 p or ^1_1 H).

How do I check whether a nuclear equation is correct?

Check both conservation laws. For example, in ^238_92 U -> ^234_90 Th + ^4_2 He: mass numbers 238 = 234 + 4, and atomic numbers 92 = 90 + 2. If either sum fails, the equation is wrong or something is missing.

What do I do when a nuclear equation has an unknown particle?

Find what is missing by difference. If the equation ^222_86 Rn -> ^218_84 Po + ?, subtract: mass 222 - 218 = 4, charge 86 - 84 = 2. The missing particle is ^4_2 He, an alpha particle. Mass numbers tell you its size; atomic numbers tell you its charge.

What symbols do I write for alpha, beta, and gamma in equations?

It depends on the decay mode. An alpha particle is ^4_2 He, a beta-minus particle is ^0_-1 e, a positron is ^0_+1 e, and a gamma ray is written as ^0_0 gamma (or simply the Greek letter). In each case the '4' and '2' style numbers are chosen so that A and Z balance across the arrow.