💡 Gamma
A gamma ray carries energy away, and both its numbers are 0. Cobalt-60 (atomic number 27) can give out a gamma ray after a decay:
6027Co → 6027Co + 00γ
Radioactivity » Balancing Nuclear Equations
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 7.8
- How to write a nuclear equation with mass numbers and atomic numbers
- The symbols for an alpha particle, a beta particle, a gamma ray and a neutron
- How to check that mass and charge are balanced
- How to find a missing number or a missing particle
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Unlock This CourseWriting a nuclear decay as an equation
In Nuclear Decay you worked out what happens to a nucleus in words. Now you will write it as an equation. Each nucleus is written with its symbol, its mass number at the top and its atomic number at the bottom. For example, 23892U means uranium with mass number 238 and 92 protons.
A nuclear equation has the nucleus before the decay on the left, then an arrow, then the new nucleus and the radiation on the right. The equation is balanced when two totals match on both sides:
- the top numbers (mass numbers) add up to the same total on both sides;
- the bottom numbers (the charge, shown by the atomic number) add up to the same total on both sides.
Key terms:
- Nuclear equation: a way of showing a nuclear change using symbols, with the nucleus before on the left and the products on the right.
- Balanced: the mass numbers match on both sides and the charges match on both sides.
The four symbols to learn
| Radiation | Symbol | Top number | Bottom number |
|---|---|---|---|
| Alpha particle | 42He or 42α | 4 | 2 |
| Beta particle | 0−1e or 0−1β | 0 | −1 |
| Gamma ray | 00γ | 0 | 0 |
| Neutron | 10n | 1 | 0 |
Check each one makes sense. An alpha particle has 4 particles, 2 of them protons. A beta particle is an electron, so it has almost no mass (top number 0) and a charge of −1. A gamma ray is energy only, so both numbers are 0. A neutron has mass 1 and no charge.
Alpha emission
Use the alpha rules from Nuclear Decay, then check that both rows balance.
Worked example
Polonium-210 (atomic number 84) emits an alpha particle and becomes lead (atomic number 82).
21084Po → 20682Pb + 42He
Top numbers: 210 on the left. 206 + 4 = 210 on the right. Balanced.
Bottom numbers: 84 on the left. 82 + 2 = 84 on the right. Balanced.
Beta emission
Use the beta rules from Nuclear Decay. The electron has a charge of −1, so the new nucleus must have a charge one higher to keep the total the same.
Worked example
Strontium-90 (atomic number 38) emits a beta particle and becomes yttrium (atomic number 39).
9038Sr → 9039Y + 0−1e
Top numbers: 90 on the left. 90 + 0 = 90 on the right. Balanced.
Bottom numbers: 38 on the left. 39 + (−1) = 38 on the right. Balanced.
Gamma and neutron emission
🔄 Neutron
A neutron has mass 1 and no charge. Helium-5 can give out a neutron:
52He → 42He + 10n
Check the neutron one: top 5 = 4 + 1, bottom 2 = 2 + 0. Both balance.
Finding a missing number
Add up one side, then take away the known number on the other side.
Worked example
Iodine-131 emits a beta particle: 13153I → 131?Xe + 0−1e
Bottom numbers: 53 = ? + (−1)
? = 53 + 1 = 54
The xenon nucleus has atomic number 54.
Identifying a missing particle
Work out the top and bottom numbers of the missing particle, then match them to the table of symbols.
Worked example
Radon-222 decays: 22286Rn → 21884Po + X
Top number of X: 222 − 218 = 4
Bottom number of X: 86 − 84 = 2
A particle with top number 4 and bottom number 2 is an alpha particle, 42He. So X is an alpha particle.
Common mistakes
Common mistakes
- Adding the mass number and atomic number together. Always do the top row and the bottom row separately.
- Forgetting the −1 on a beta particle. Then the bottom numbers will not balance.
- Writing the alpha particle as 44He. The bottom number is 2, because it has 2 protons.
- Not checking the answer. Add up both rows on each side at the end. It takes ten seconds.
Exam-style question
(a) Thorium-232 decays by emitting a particle X: 23290Th → 22888Ra + X. Identify X and give its symbol. [2 marks]
(b) Cobalt-60 has atomic number 27. It decays by beta emission into nickel-60. Complete the equation by finding the atomic number of nickel: 6027Co → 60?Ni + 0−1e [2 marks]
(c) State what must be true about the numbers on the two sides of any balanced nuclear equation. [2 marks]
Model answer
(a) X has top number 232 − 228 = 4 and bottom number 90 − 88 = 2 (1). It is an alpha particle, 42He (1).
(b) 27 = ? + (−1) (1), so ? = 28 (1).
(c) The mass numbers (top numbers) add up to the same total on both sides (1). The charges (bottom numbers) add up to the same total on both sides (1).
Exam tip
In part (b), write "27 = ? − 1" or similar before you give the answer. If the final number is wrong, showing the sum can still earn the method mark.