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Practise 35 exam-style questions on Nuclear equations for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Phosphorus-32 is used in medicine. It decays by beta emission to form sulfur-32. Sulfur has an atomic number of 16.
What is the atomic number of phosphorus?
15
Beta decay increases the atomic number by one, so the nucleus before the decay had an atomic number one less than sulfur: 16 − 1 = 15.
15
Technetium-99m is used in medical imaging. It is an excited form of the technetium-99 nucleus. It emits a gamma ray:
⁹⁹₄₃Tc → ⁹⁹₄₃Tc + γ
Explain why the mass number and the atomic number do not change.
A gamma ray is electromagnetic radiation. It has no mass, so the number of nucleons, and so the mass number, stays the same. It has no charge, so the number of protons, and so the atomic number, stays the same. The nucleus just loses energy.
Gamma emission usually happens after another decay has left the nucleus with extra energy. The nucleus keeps all its protons and neutrons, so it is the same isotope of the same element afterwards.
a gamma ray has no mass / contains no protons or neutrons, so the mass number does not change
allow gamma is electromagnetic radiation, not a particle
a gamma ray has no charge, so the atomic number does not change
The equation shows the decay of radium-226:
²²⁶₈₈Ra → ²²²₈₆Rn + a particle with mass number P and atomic number Q
Determine the values of P and Q, and name the particle.
P = 226 − 222 = 4
Q = 88 − 86 = 2
A particle with mass number 4 and atomic number 2 is an alpha particle.
Balance the top numbers and the bottom numbers separately. A particle with mass number 4 and atomic number 2 is two protons and two neutrons, which is an alpha particle (a helium nucleus).
P = 4
correct answer with no working scores full marks
Q = 2
alpha particle
allow helium nucleus; allow ecf from incorrect P and Q
Americium-241 is used in smoke detectors. It is produced when plutonium-241 decays by beta emission. Americium-241 then decays by alpha emission to form neptunium.
Plutonium has atomic number 94. The symbols are Pu for plutonium, Am for americium and Np for neptunium.
Write balanced nuclear equations for both decays.
²⁴¹₉₄Pu → ²⁴¹₉₅Am + ⁰₋₁e
²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He
Work out americium's numbers first from the beta decay (same mass number, atomic number up by one), then use them as the starting point for the alpha decay (mass number down by 4, atomic number down by 2). Check that each equation balances.
²⁴¹₉₄Pu → ²⁴¹₉₅Am: americium with mass number 241 and atomic number 95
allow any clear notation, e.g. 241/94 Pu → 241/95 Am + 0/−1 e
+ ⁰₋₁e (so the first equation is ²⁴¹₉₄Pu → ²⁴¹₉₅Am + ⁰₋₁e)
allow ⁰₋₁β
²⁴¹₉₅Am → ²³⁷₉₃Np: neptunium with mass number 237 and atomic number 93
allow ecf from an incorrect atomic number for americium
+ ⁴₂He (so the second equation is ²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He)
allow ⁴₂α
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