Practise Nuclear equations. 15 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
How to write and balance nuclear equations for alpha and beta decay, and how alpha, beta and gamma emission change the mass and charge of a nucleus. Expect 'complete the equation' questions and questions asking you to identify the type of decay from the numbers.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Recall the symbols for alpha and betaAlpha is \({}^{4}_{2}\mathrm{He}\) and beta is \({}^{0}_{-1}\mathrm{e}\).
5
State how alpha decay changes a nucleusThe mass number falls by 4 and the atomic number falls by 2.
5
State how beta decay changes a nucleusThe mass number stays the same and the atomic number goes up by 1.
5
Explain why gamma emission changes neither numberGamma is electromagnetic radiation with no mass and no charge.
6
Complete a nuclear equationMake the mass numbers and the atomic numbers balance on both sides of the arrow.
7
Identify the decay type from an equationCompare the mass and atomic numbers before and after to see what was emitted.
8
Work through a chain of decaysApply several alpha and beta decays in turn to find the final nucleus.
Notes
Symbols in nuclear equations
Each nucleus is written as \({}^{A}_{Z}\mathrm{X}\): A is the mass number and Z is the atomic number (its charge).
Alpha particle: \({}^{4}_{2}\mathrm{He}\) (sometimes written \({}^{4}_{2}\alpha\)).
Beta particle: \({}^{0}_{-1}\mathrm{e}\) (sometimes written \({}^{0}_{-1}\beta\)). Mass number 0, charge −1.
The mass numbers (top numbers) add up to the same total on both sides of the arrow.
The atomic numbers (bottom numbers) add up to the same total on both sides.
If the atomic number changes, the element changes. Use the new atomic number to find the new element.
What each type of decay does
Alpha decay: mass number down by 4, atomic number down by 2. Both the mass and the charge of the nucleus decrease. Example: \({}^{226}_{88}\mathrm{Ra} \rightarrow {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\mathrm{He}\)
Beta decay: mass number unchanged, atomic number up by 1. The mass does not change but the charge increases, because a neutron has become a proton. Example: \({}^{90}_{38}\mathrm{Sr} \rightarrow {}^{90}_{39}\mathrm{Y} + {}^{0}_{-1}\mathrm{e}\)
Gamma emission: no change to the mass or the charge of the nucleus. The nucleus just gets rid of energy, often straight after an alpha or beta decay.
Neutron emission: the mass number falls by 1 and the atomic number stays the same, so the nucleus becomes a different isotope of the same element. grade 7+
Cheatsheet
Alpha \({}^{4}_{2}\mathrm{He}\): mass number − 4, atomic number − 2
Beta \({}^{0}_{-1}\mathrm{e}\): mass number unchanged, atomic number + 1
Gamma \({}^{0}_{0}\gamma\): no change to mass number or atomic number
Neutron \({}^{1}_{0}\mathrm{n}\): mass number − 1, atomic number unchanged
Top numbers balance; bottom numbers balance
New atomic number = new element
How to answer each type of question
Describe how a decay changes the nucleus
2 marks5
For each decay, say what happens to the mass (mass number).
Then say what happens to the charge (atomic number).
Example. Compare the effect on a nucleus of emitting an alpha particle and emitting a gamma ray.
Show the model answer
Alpha: the mass number decreases by 4 and the atomic number (charge) decreases by 2 (1). Gamma: there is no change to the mass or the charge of the nucleus (1).
Complete a nuclear equation
2 marks6
Write in the full symbol for the particle emitted, with both numbers.
Balance the top numbers.
Balance the bottom numbers (remember beta counts as −1).
If asked, name the new element from its atomic number.
Example. Polonium-210 decays by emitting an alpha particle. The equation for the decay is \({}^{210}_{84}\mathrm{Po} \rightarrow {}^{a}_{b}\mathrm{Pb} + {}^{4}_{2}\mathrm{He}\) Determine the values of a and b.
Show the model answer
a = 210 − 4 = 206 (1) b = 84 − 2 = 82 (1)
Identify the type of decay
2 marks7
Find the change in mass number and in atomic number.
Work out the mass number and charge of the missing particle.
Name it and give the numbers as your reason.
Example. Thorium-234 decays to protactinium-234: \({}^{234}_{90}\mathrm{Th} \rightarrow {}^{234}_{91}\mathrm{Pa} + \mathrm{X}\) Identify X. Explain how you know.
Show the model answer
X is a beta particle, \({}^{0}_{-1}\mathrm{e}\) (1). The mass number stays at 234 but the atomic number goes up by 1, so X has mass number 0 and charge −1 (1).
Work through a chain of decays
3 marks8
Apply each decay in the order given, writing down the new numbers each time.
Alpha: − 4 and − 2. Beta: no change and + 1.
Check whether the final atomic number matches the starting element.
Example. A uranium-238 nucleus, \({}^{238}_{92}\mathrm{U}\), emits an alpha particle. The nucleus formed then emits two beta particles, one after the other. Determine the mass number and atomic number of the final nucleus.
Show the model answer
After the alpha decay: mass number 234, atomic number 90 (1). After two beta decays: mass number still 234 (1), atomic number 90 + 2 = 92 (1). (The final nucleus is uranium-234, an isotope of the original element.)
Shortcuts and memory tricks
Alpha: 'minus 4, minus 2'. Beta: 'same, plus 1'. Gamma: 'no change'.
Check every equation with two quick sums: the top numbers on each side, then the bottom numbers.
The −1 on the beta particle is why the new nucleus goes UP by 1: the bottom numbers must still add up.
Where marks are lost
Writing the beta particle as \({}^{0}_{1}\mathrm{e}\) or \({}^{1}_{0}\mathrm{e}\). It is \({}^{0}_{-1}\mathrm{e}\).
Taking 2 off the mass number in alpha decay. It is 4 off the mass number and 2 off the atomic number.
Keeping the same element symbol after alpha or beta decay. The atomic number has changed, so the element has too.
Thinking beta decay lowers the atomic number. It raises it by 1.
Exam technique
Write both numbers for every particle in the equation, even if the question only asks for one.
If a periodic table or list of elements is given, use the new atomic number to choose the symbol.
When asked to 'describe' the change, use words too: e.g. 'the mass stays the same and the charge increases'.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
What happens to the mass number and to the atomic number of a nucleus when it emits an alpha particle?
Mass number falls by 4; atomic number falls by 2.
Explain how the equation shows that emitting a neutron does not change the charge of the nucleus.
The atomic number stays at 2, because the neutron's atomic number is 0.
In a balanced nuclear equation, what must be true about the mass numbers on each side of the equation?
The total mass number is the same on both sides.
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) Which symbol represents an alpha particle? Tick (✓) one box.[1]
\({}^{4}_{2}\mathrm{He}\)
\({}^{0}_{-1}\mathrm{e}\)
\({}^{1}_{0}\mathrm{n}\)
\(\gamma\)
(b) Which symbol represents a beta particle? Tick (✓) one box.[1]
\({}^{4}_{2}\mathrm{He}\)
\({}^{0}_{-1}\mathrm{e}\)
\({}^{1}_{0}\mathrm{n}\)
\(\gamma\)
(c) What happens to the mass number and to the atomic number of a nucleus when it emits an alpha particle?[2]
(d) What happens to the mass number and to the atomic number of a nucleus when it emits a gamma ray?[1]
Show the answer and mark scheme
(a)Answer: \({}^{4}_{2}\mathrm{He}\)
(b)Answer: \({}^{0}_{-1}\mathrm{e}\)
(c)Answer: Mass number falls by 4; atomic number falls by 2.
the mass number decreases by 4
the atomic number decreases by 2
(d)Answer: Neither changes.
neither changes / both stay the same
Question 2Medium7 marks
Radium-226 decays by emitting an alpha particle. The equation for this decay is: \({}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{y}\mathrm{Rn} + {}^{4}_{2}\mathrm{He}\)
(a) Determine the values of x and y in the equation.[2]
(b) Strontium-90, \({}^{90}_{38}\mathrm{Sr}\), decays by emitting a beta particle. The nucleus formed is an isotope of yttrium (Y). Write a balanced nuclear equation for this decay.[3]
(c) Explain why the mass number does not change when a nucleus emits a beta particle.[2]
beta particle shown as \({}^{0}_{-1}\mathrm{e}\) on the right-hand side
yttrium with mass number 90
yttrium with atomic number 39
(c)Answer: A neutron turns into a proton, so the total number of protons and neutrons is unchanged.
a neutron in the nucleus changes into a proton (and a high-speed electron is emitted)
so the total number of protons and neutrons stays the same
Question 3Hard7 marks
Some very unstable nuclei decay by emitting a neutron. A helium-5 nucleus decays in this way: \({}^{5}_{2}\mathrm{He} \rightarrow {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}\)
(a) Explain how the equation shows that emitting a neutron does not change the charge of the nucleus.[1]
(b) Explain why neutron emission always produces an isotope of the same element.[2]
(c) Compare the changes to the mass number and the atomic number of a nucleus caused by alpha decay, beta decay, gamma emission and neutron emission.[4]
Show the answer and mark scheme
(a)Answer: The atomic number stays at 2, because the neutron's atomic number is 0.
the atomic number (number of protons) stays at 2 / the neutron has an atomic number (charge) of 0
(b)Answer: The number of protons is unchanged (same element) but there is one fewer neutron (different isotope).
the number of protons (atomic number) does not change, so it is the same element
the number of neutrons (mass number) decreases by 1, so it is a different isotope