Nuclear equations and conservation

Key idea: Nuclear equations conserve charge and nucleon number. Mass can change, but total mass–energy is conserved, with released energy given by Δmc².

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  • GCE A-Level H1 Physics 2027
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In 1919 Rutherford fired alpha particles into nitrogen gas and found protons coming out. He had turned nitrogen into oxygen, the first artificial nuclear reaction. A nuclear equation records what happens in a reaction like this, and two simple counts, of nucleons and of charge, let you work out any particle that is missing.

Nuclear equations

In a nuclear equation each particle is written as X with nucleon number A and proton number Z. The top number is the nucleon number. The bottom number is the charge in units of e: for a nucleus that is its proton number, but for an electron it is -1.

ParticleSymbolNucleon numberCharge in units of e
proton¹₁p or ¹₁H1+ 1
neutron¹₀n10
alpha particle⁴₂He or ⁴₂α4+ 2
beta particle⁰₋₁e or ⁰₋₁β0-1
gamma radiation⁰₀γ00

In every nuclear reaction:

  • the total nucleon number is the same on both sides;
  • the total charge is the same on both sides.

Balance the two separately. A number in front of a particle multiplies both its nucleon number and its charge.

Worked example 1

Rutherford’s reaction

Problem

Identify the particle X in Rutherford’s reaction:

¹⁴₇N + ⁴₂He → ¹⁷₈O + X
Show full solution
  1. Nucleon number

    Method

    Balance the top numbers.

    Reason

    Nucleon number is conserved.

    Working

    Answer

    14 + 4 = 17 + A, so A = 1

  2. Charge

    Method

    Balance the bottom numbers.

    Reason

    Charge is conserved, and it tells a proton from a neutron.

    Working

    Answer

    7 + 2 = 8 + Z, so Z = 1

  3. The particle

    Working

    Answer

    X is ¹₁H, a proton.

Decay equations

In radioactive decay a single parent nucleus changes into a daughter nucleus, and the same two counts apply.

  • Alpha decay: the nucleus loses 2 protons and 2 neutrons, so A falls by 4 and Z by 2.
  • Beta decay: a neutron in the nucleus changes into a proton and an electron, which is emitted. A stays the same and Z rises by 1. An antineutrino is also emitted; it has no charge and no nucleon number, so it does not affect the balance.
  • Gamma emission: the nucleus loses energy only, so A and Z do not change. It often follows alpha or beta decay.

Guided practice 2

What leaves a plutonium nucleus?

About 3 min

Problem

Find the nucleon number and charge of the particle X. Use the particle table to name it when you compare your answer:

²³⁹₉₄Pu → ²³⁵₉₂U + X

Find both numbers before naming the particle

Hints

Hint 1

two differences

Subtract the daughter’s top number from the parent’s, then do the same for the bottom numbers.

Show solution step by step
  1. The two numbers

    Method

    Find the differences between parent and daughter.

    Reason

    The emitted particle carries them away.

    Working

    Answer

    A = 239-235 = 4 and Z = 94-92 = 2

  2. The particle

    Working

    Answer

    X is ⁴₂He, an alpha particle.

Check your understanding 1 Carbon-14 decay

Carbon-14, ¹⁴₆C, decays by beta emission. Write the equation and name the daughter nucleus.

Show answer

¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν bar ₑ. The electron antineutrino ν bar ₑ carries zero nucleon number and zero charge. The nucleon number stays 14 and the proton number rises to 7, so the daughter is nitrogen-14. Check: 6 = 7 + (-1).

Conservation of mass–energy

Nucleon number and charge are conserved in every nuclear process. So is mass–energy, but mass on its own is not. In a reaction that releases energy, the total mass of the products is slightly less than the total mass before. The missing mass Δ m appears as energy Δ E = Δ m c², mostly as kinetic energy of the products and as gamma radiation.

A balanced equation shows that nucleon number and charge are conserved. It does not show how much energy is released; for that you need the masses of the particles.

Try it yourself 3

Neutrons from fission

Minimal support

Problem

A uranium-235 nucleus absorbs a neutron and splits into two smaller nuclei. Find x, check that charge is conserved, and say what must happen to the total mass if energy is released.

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x ¹₀n

Count the incoming neutron too

Hints

Hint 1

nucleons

Include the neutron on the left. The x in front of the neutron on the right multiplies its nucleon number.

Hint 2

energy

Energy released comes from a change in mass.

Show solution step by step
  1. Nucleon number

    Method

    Balance the top numbers.

    Reason

    Each neutron contributes one nucleon.

    Working

    Answer

    235 + 1 = 141 + 92 + x, so x = 3

  2. Charge

    Method

    Balance the bottom numbers.

    Reason

    Neutrons have no charge, whatever their number.

    Working

    Answer

    92 + 0 = 56 + 36 + 0, so charge is conserved.

  3. Mass

    Method

    Apply conservation of mass–energy.

    Reason

    Energy released must come from somewhere.

    Working

    Answer

    The total mass of the products is less than the total mass of the uranium nucleus and the neutron. The difference is released as energy, Δ E = Δ m c².

Common mistakes

  • Balancing only the nucleon numbers. Charge must balance too, and it is what tells a proton from a neutron.
  • Writing the charge of an electron as 0 or + 1. It is -1.
  • Forgetting the incoming neutron in a fission equation, or forgetting to multiply by the number of neutrons released.
  • Saying mass is conserved in a nuclear reaction. Mass–energy is conserved; the mass can change.
  • Thinking that beta decay leaves the nucleon number changed. A neutron becomes a proton, so A stays the same.

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