Conservation of charge and nucleon number

8 minAS Level11.1.6

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can state the two conservation rules for nuclear processes, check a decay equation with them, and use them to work out the numbers for an emitted particle or a daughter nuclide.

Definition

Conservation in nuclear processes

In any nuclear process:

the total charge of the particles before equals the total charge after, and
the total nucleon number before equals the total nucleon number after.

Because charge and nucleon number are written into nuclide notation as $Z$ and $A$ , these become two arithmetic rules: the $Z$ values must add up on both sides, and the $A$ values must add up on both sides.

Why the two sums must match

Charge is conserved in every interaction, nuclear or not; a nucleus cannot create or destroy charge. Since a nuclide's charge is $+Ze$ , conserving charge means conserving the sum of the $Z$ values.

Nucleon number is conserved because a nuclear decay rearranges nucleons rather than making them. In beta-minus decay a neutron turns into a proton, so the count of nucleons is unchanged even though the split between protons and neutrons changes.

Check uranium-238 alpha decay term by term:

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

	left	right
nucleon number $A$	238	$234 + 4 = 238$
proton number $Z$	92	$90 + 2 = 92$

Both balance, so the equation is possible. Now beta-minus decay of carbon-14:

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

Nucleon number: $14 = 14 + 0$ . Charge: $6 = 7 + (-1)$ . The electron's $Z = -1$ is what allows the proton number to rise by one while charge is still conserved.

Balancing, not guessing

Treat a decay equation like a set of scales with two independent balances, one for $A$ and one for $Z$ . Because the two balances are independent, you can often find the unknown particle from $A$ alone and then confirm which particle it is from $Z$ .

Take $^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + \text{?}$ .

The nucleon numbers need $226 = 222 + A$ , so $A = 4$ . The proton numbers need $88 = 86 + Z$ , so $Z = 2$ . A particle with $A = 4$ and $Z = 2$ is an alpha particle, so the decay is alpha emission. If the arithmetic had given $A = 0$ and $Z = -1$ , the particle would have been a beta-minus particle instead.

This is the standard exam technique: balance $A$ , balance $Z$ , then name the particle.

Worked example: Finding the daughter nucleus

A nucleus of $^{238}_{92}\text{U}$ emits an alpha particle. Use the conservation rules to write the decay equation.

Have a go first?

Quick check

CalculateQuestion 1 · Apply

Multiple choiceQuestion 2 · Recall

Answer the 3 questions above to continue.