Nuclear Equations

A nuclear equation is a way of writing down what happens to a nucleus when it decays. It shows the original nucleus on the left, and what it turns into on the right, using the same mass-number and atomic-number notation you already know.

Every particle in a nuclear equation is written in the form:

${}_Z^A\text{X}$

where A is the mass number (top, protons + neutrons) and Z is the atomic number (bottom, number of protons). The single rule that governs all of these equations is that both totals must balance.

In any nuclear equation, the total mass number must be the same on both sides, and the total atomic number must be the same on both sides.

The specification asks you to write balanced equations for single alpha and beta decays. Importantly, you are not required to identify the daughter element by name or symbol; the marks are for balancing the mass and atomic numbers correctly.

Two decay-particle symbols are set by the specification and must be memorised exactly.

• An alpha particle is written ${}_2^4\text{He}$ (it is a helium nucleus: 2 protons, 2 neutrons).
• A beta particle is written ${}_{-1}^0\text{e}$ (a high-speed electron from the nucleus).

The beta symbol looks strange at first, so it is worth unpacking:

• The top number is 0, because an electron has a negligible mass number.
• The bottom number is −1, because an electron carries a charge of −1, the opposite of a proton's +1.

Gamma radiation can be written ${}_0^0\gamma$: zero mass number and zero atomic number, since it is electromagnetic radiation carrying neither mass nor charge. This is why a gamma ray never changes the numbers in an equation.

In alpha decay the nucleus emits an alpha particle, ${}_2^4\text{He}$. Since the alpha particle takes away 2 protons and 2 neutrons, the effect on the original nucleus is fixed.

Alpha decay decreases the mass number by 4 and the atomic number by 2.

Worked example — a radium-226 nucleus, ${}_{88}^{226}\text{Ra}$, undergoes alpha decay. Write the balanced equation.

Start by subtracting the alpha particle's numbers from the original nucleus:

• New mass number = $226-4=222$
• New atomic number = $88-2=86$

${}_{88}^{226}\text{Ra}\to {}_{86}^{222}\text{Rn}+{}_2^4\text{He}$

Now check it balances:

• Mass numbers: $222+4=226$ ✓ (matches the left side)
• Atomic numbers: $86+2=88$ ✓ (matches the left side)

You would still gain full marks writing the new nucleus as ${}_{86}^{222}\text{X}$, because identifying it as radon is not required.

In beta decay a neutron in the nucleus turns into a proton, and a high-speed electron, the beta particle ${}_{-1}^0\text{e}$, is emitted.

Because a neutron becomes a proton, the total number of nucleons (protons + neutrons) does not change, so the mass number stays the same. But there is now one extra proton, so the atomic number rises by 1.

Beta decay leaves the mass number unchanged and increases the atomic number by 1.

Worked example — a carbon-14 nucleus, ${}_6^{14}\text{C}$, undergoes beta decay. Write the balanced equation.

• Mass number is unchanged: stays 14
• Atomic number increases by 1: $6+1=7$

${}_6^{14}\text{C}\to {}_7^{14}\text{N}+{}_{-1}^0\text{e}$

Check it balances:

• Mass numbers: $14+0=14$ ✓
• Atomic numbers: $7+(-1)=6$ ✓

The −1 on the beta particle is essential to the balance. It is exactly what allows the atomic number to increase from 6 to 7 while the two sides still add up.

A nucleus often emits gamma radiation after alpha or beta decay, to lose surplus energy.

Gamma emission changes neither the mass number nor the atomic number. Gamma is energy only, with no mass and no charge.

For that reason gamma does not turn one nucleus into another; it simply carries energy away.

The reliable method for any decay equation is to balance the numbers rather than memorise outcomes:

1. Add up the mass numbers (top) on the left; the right side must total the same.
2. Add up the atomic numbers (bottom) on the left; the right side must total the same.
3. Use those two totals to fill in any missing number.

Worked example — a nucleus decays as ${}_{86}^{219}\text{Rn}\to {}_Z^A\text{X}+{}_2^4\text{He}$. Find A and Z.

• Mass number: $A+4=219$, so $A=219-4=215$
• Atomic number: $Z+2=86$, so $Z=86-2=84$

The missing nucleus is ${}_{84}^{215}\text{X}$. Balancing the two totals gives the answer without needing to know which element it is.

1. Writing the beta particle without the −1

The beta particle is ${}_{-1}^0\text{e}$. Leaving off the −1, or writing +1, breaks the balance and the atomic numbers will not add up.

2. Getting alpha decay changes the wrong way round

Alpha decay decreases the mass number by 4 and the atomic number by 2. Both go down. A common slip is subtracting 2 from the mass number or 4 from the atomic number.

3. Changing the mass number in beta decay

In beta decay the mass number stays the same; only the atomic number rises by 1. A neutron becomes a proton, so the total nucleon count is unchanged.

4. Thinking gamma emission changes the nucleus

Gamma has mass number 0 and atomic number 0, so it changes neither number. It only carries energy away.

5. Not checking the equation balances

Always confirm the mass numbers on the right add up to the mass number on the left, and the same for atomic numbers. This one check catches almost every arithmetic error.