Physics 9702 · Paper 2

Decay Equations

When a nucleus decays it becomes a different nucleus. We track the change with an equation. The whole skill is keeping two totals balanced.

Two totals must balance

Whatever comes out of a decay, two totals are the same before and after.
The nucleon number $A$ (the top numbers) must add up the same on both sides.
The proton number $Z$ (the bottom numbers) must add up the same on both sides. Because $Z$ is really the charge, this is charge being conserved.
If you know the particle that is emitted, these two rules tell you what the new nucleus must be. That is all a decay equation is.

Alpha and beta decay

In alpha decay the nucleus loses 2 protons and 2 neutrons, so $A$ drops by 4 and $Z$ drops by 2:

{}^{238}_{92}\text{U} \rightarrow {}^{234}_{90}\text{Th} + {}^{4}_{2}\alpha

Check it: nucleon numbers $238 = 234 + 4$ , proton numbers $92 = 90 + 2$ . Both balance, so the new element is the one with proton number 90, thorium.
In beta-minus decay a neutron in the nucleus turns into a proton and releases an electron. The nucleon number stays the same (a neutron became a proton, still one nucleon), but the proton number goes up by 1:

{}^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{\ \ 0}_{-1}\beta + \bar{\nu}

The electron is written ${}^{\ \ 0}_{-1}\beta$ : no nucleons (top 0), charge $-1$ (bottom). That $-1$ lets the proton number rise from 6 to 7 while the charge still balances: $6 = 7 + (-1)$ .
The $\bar{\nu}$ is an antineutrino. It has no charge and no nucleons, so it does not change the balancing. Its partner, the neutrino $\nu$ , is released in beta-plus decay instead.

In beta-plus decay a proton turns into a neutron and releases a positron. The nucleon number stays the same, but the proton number goes down by 1:

{}^{12}_{\ 7}\text{N} \rightarrow {}^{12}_{\ 6}\text{C} + {}^{\ \ 0}_{+1}\beta + \nu

The positron is written ${}^{\ \ 0}_{+1}\beta$ : charge $+1$ . That lets the proton number fall from 7 to 6 while the charge balances: $7 = 6 + 1$ . A neutrino $\nu$ is released, not an antineutrino. (9702/21/M/J/25 Q7(a))
Gamma emission is the simplest. The nucleus releases a burst of energy and drops to a lower energy state. No protons or neutrons leave, so $A$ and $Z$ are both unchanged. It is the same nucleus with less energy.

Worked example

Balancing a decay

Radium-226 ( ${}^{226}_{88}\text{Ra}$ ) decays by alpha emission. What is the new nucleus?

Alpha decay removes 2 protons and 4 nucleons: $Z: 88 - 2 = 86$ , $A: 226 - 4 = 222$ .
Proton number 86 is radon (Rn), so the nucleus becomes ${}^{222}_{86}\text{Rn}$ .

Answer

{}^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\alpha

Your turn— tap to reveal the worked answer (9702/21/M/J/24 Q7)

Copper-66 ( ${}^{66}_{29}\text{Cu}$ ) decays by beta-minus emission to zinc (Zn). Write the decay equation. (9702/21/M/J/24 Q7(b))

Answer: $A$ is unchanged and $Z$ rises by 1, so zinc is ${}^{66}_{30}\text{Zn}$ :

{}^{66}_{29}\text{Cu} \rightarrow {}^{66}_{30}\text{Zn} + {}^{\ \ 0}_{-1}\beta + \bar{\nu}

Why there is a neutrino

Beta particles leave with a range of energies, from zero up to one maximum. The decay always releases the same fixed energy (the dashed line). The missing energy is taken away by the antineutrino (β⁻) or neutrino (β⁺) — which is why that extra particle must be there.

A decay always releases the same fixed energy. You would expect every beta-particle to come out with that one energy.
But beta-particles come out with a whole range of energies, from zero up to one maximum. Some energy is missing.
The missing energy is taken away by another particle with no charge: the antineutrino (in beta-minus) or the neutrino (in beta-plus). This is why that extra particle must be in the equation.
This also lets you read a graph. A continuous range of particle energies means beta decay. A single fixed energy means alpha decay. (9702/23/M/J/24 Q6) (9702/22/M/J/25 Q7)
The quantities conserved in any decay are the nucleon number, the proton number (charge), momentum and mass–energy. (9702/22/F/M/25 Q7(c))

Common mistake

Get the partner right: beta-minus gives an antineutrino, beta-plus gives a neutrino. And when a question shows a range of energies and asks which decay it is, a range means beta, not alpha. Alpha-particles all leave with the same energy.

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