10 questions that get harder as you go. No calculator. Write your working on paper, pick A–E, then open the solution. Q1–3 are warm-ups, Q4–6 a step up, Q7–10 ESAT standard.
Q1 · Warm-up
Which type of radiation is the most strongly ionising?
A Gamma · B Beta · C Alpha · D X-rays · E They are all equally ionising
Show solution
C. Alpha particles are big and have a +2 charge, so they ionise strongly. That’s also why they’re stopped by paper or a few cm of air.
Q2 · Warm-up
An isotope of carbon is written in nuclide notation as 146C. How many neutrons are in its nucleus?
A 6 · B 14 · C 20 · D 8 · E 7
Show solution
D. The top number is the mass number (protons + neutrons) and the bottom number is the atomic number (protons). Neutrons = 14 − 6 = 8.
💡 Every carbon atom has 6 protons; that’s what makes it carbon. Different isotopes of carbon (like carbon-12 and carbon-14) just have different numbers of neutrons.
Q3 · Warm-up
Radium-226 (atomic number 88) decays by alpha emission. What are the mass number and atomic number of the new nucleus?
A 222, 86 · B 226, 87 · C 222, 90 · D 224, 86 · E 226, 89
Show solution
A. An alpha particle is 2 protons + 2 neutrons, so mass number − 4 and atomic number − 2: 226 − 4 = 222 and 88 − 2 = 86.
Q4 · Step up
What happens to a nucleus when it emits a beta-minus particle?
A Mass number − 4, atomic number − 2 · B Mass number unchanged, atomic number + 1 · C Mass number unchanged, atomic number − 1 · D Mass number + 1, atomic number unchanged · E Neither changes
Show solution
B. A neutron turns into a proton and an electron, and the electron is fired out. One more proton means atomic number + 1. The total number of particles in the nucleus stays the same, so the mass number doesn’t change. (E describes gamma.)
Q5 · Step up
An isotope has a half-life of 6 hours. What fraction of it is left undecayed after 24 hours?
A 1/4 · B 1/8 · C 1/24 · D 1/16 · E 1/32
Show solution
D. 24 ÷ 6 = 4 half-lives. Halve four times: 1 → ½ → ¼ → ⅛ → 1/16.
Q6 · Step up
The graph shows how the count rate from a radioactive source (background already removed) falls over 12 hours. Use the graph to find the half-life, then work out the count rate after 15 hours.
A 50 counts/min · B 100 counts/min · C 0 counts/min · D 12.5 counts/min · E 25 counts/min
Show solution
E. Read the half-life off the graph: the count rate halves from 800 to 400 in 3 hours (and from 400 to 200 in the next 3). So the half-life is 3 hours. 15 hours is 5 half-lives: 800 → 400 → 200 → 100 → 50 → 25 counts/min.
💡 C is the trap: a decay curve never actually reaches zero. It halves again and again. A stops at 12 hours, the end of the graph.
Q7 · ESAT standard
A detector near a source reads 260 counts per minute. The background count is 20 counts per minute. What will the detector read after two half-lives?
A 80 counts/min · B 60 counts/min · C 65 counts/min · D 85 counts/min · E 120 counts/min
Show solution
A. Only the source decays, not the background. Source alone = 260 − 20 = 240. After two half-lives: 240 → 120 → 60. The detector still picks up background, so it reads 60 + 20 = 80.
💡 B forgets to add the background back. C halves the background too.
Q8 · ESAT standard
Uranium-238 (atomic number 92) decays through a chain to lead-206 (atomic number 82). How many alpha and beta-minus decays happen in the chain?
A 6 alpha, 8 beta · B 8 alpha, 10 beta · C 8 alpha, 6 beta · D 10 alpha, 8 beta · E 8 alpha, 2 beta
Show solution
C. Only alpha changes the mass number. It drops by 238 − 206 = 32, so that’s 32 ÷ 4 = 8 alphas.
8 alphas lower the atomic number by 16: 92 − 16 = 76. But lead is 82, so the atomic number must go back up by 6. Each beta-minus adds 1, so there are 6 betas.
Q9 · ESAT standard
A narrow beam containing alpha, beta and gamma radiation passes, in a vacuum, between two parallel metal plates. The top plate is positive and the bottom plate is negative. Which statements are correct?
I The beta particles bend towards the positive plate.
II The gamma rays pass between the plates without bending.
III The alpha particles bend more than the beta particles, because they have a bigger charge.
A I only · B I and II only · C II and III only · D I, II and III · E II only
Show solution
B. Beta particles are electrons, so they are negative and are pulled towards the positive plate (I ✓). Gamma rays are electromagnetic waves with no charge, so they go straight through (II ✓). Alpha particles bend the other way, towards the negative plate, but much less than beta: an alpha particle has twice the charge but about 7000 times the mass of an electron, so it is far harder to push off course (III ✗).
💡 The same pattern holds in a magnetic field: alpha and beta curve opposite ways, beta curves much more, and gamma is undeflected. Penetration runs the other way round from ionisation: alpha is stopped by paper, beta by a few mm of aluminium, and gamma is only reduced by thick lead or concrete.
Q10 · ESAT standard
Isotope X has a half-life of 2 days and isotope Y has a half-life of 4 days. At the start there are four times as many X atoms as Y atoms. After how long are the numbers of X and Y atoms equal?
A 2 days · B 4 days · C 6 days · D 8 days · E They are never equal
Show solution
D. Quickest way is a table. Start with X = 4 and Y = 1.
After 4 days: X has had 2 half-lives → 1. Y has had 1 → ½.
After 8 days: X has had 4 half-lives → ¼. Y has had 2 → ¼. Equal ✓
Algebra version: 4 × (½)t/2 = (½)t/4. Write 4 = 2², so 2 − t/2 = −t/4, giving t = 8 days.
🔥 Challenge: full ESAT level
Five harder questions with up to 8 options, answers in symbols and “which statements are correct?” questions, just like the real test. Aim for 90 seconds each.
C1
A radioactive sample is left for n half-lives. What fraction of the original nuclei has decayed?
A (½)ⁿ · B 1 − (½)ⁿ · C n/2 · D 1 − n/2 · E 2ⁿ · F 1/(2n) · G 1 − 2ⁿ · H (½)ⁿ⁻¹
Show solution
B. After each half-life, half remain, so after n half-lives the fraction left is (½)ⁿ. The fraction decayed is everything else: 1 − (½)ⁿ.
💡 Test with n = 1: B gives ½ ✓. With n = 2: B gives ¾ ✓. A is the trap: that’s what’s left, not what has decayed.
C2
A nucleus emits a gamma ray. Which statements are correct?
I Its atomic number is unchanged.
II Its mass number is unchanged.
III The nucleus loses energy.
A none · B I only · C II only · D III only · E I and II only · F I and III only · G II and III only · H I, II and III
Show solution
H. A gamma ray is an electromagnetic wave with no charge and no mass, so the numbers of protons and neutrons don’t change (I ✓, II ✓). It carries energy away, so the nucleus drops to a lower energy state (III ✓).
C3
An archaeologist finds a wooden tool that has 1/8 of the carbon-14 activity of living wood. The half-life of carbon-14 is 5700 years. Roughly how old is the tool?
A 713 years · B 1900 years · C 5700 years · D 11 400 years · E 17 100 years · F 22 800 years · G 39 900 years · H 45 600 years
Show solution
E. 1 → ½ → ¼ → ⅛ is 3 half-lives. 3 × 5700 = 17 100 years.
💡 H is the trap: 8 × 5700 counts eight half-lives instead of three. A divides by 8 instead.
C4
A nucleus X emits an alpha particle, and then the new nucleus emits two beta-minus particles. How does the final nucleus compare with X?
A Same element, mass number 4 less · B Same element, mass number 2 less · C Atomic number 2 less, mass number 4 less · D Atomic number 2 more, same mass number · E Same element, same mass number · F Atomic number 1 less, mass number 4 less · G Atomic number 4 less, mass number 2 less · H Atomic number 2 less, same mass number
Show solution
A. Alpha: mass number − 4, atomic number − 2. Each beta-minus: atomic number + 1, mass number unchanged. Overall: mass number − 4, atomic number − 2 + 2 = 0.
Same atomic number means the same element, so the final nucleus is an isotope of X, 4 mass units lighter.
C5
A sample contains two isotopes. At the start, isotope P (half-life 1 hour) gives 800 counts per second and isotope Q (half-life 3 hours) gives 100 counts per second. Ignoring background, what is the total count rate after 6 hours?
A 12.5 counts/s · B 25 counts/s · C 56.25 counts/s · D 75 counts/s · E 112.5 counts/s · F 225 counts/s · G 37.5 counts/s · H 450 counts/s
Show solution
G. Treat each isotope separately, then add.
P: 6 hours = 6 half-lives, so 800 ÷ 2⁶ = 800 ÷ 64 = 12.5 counts/s.
Q: 6 hours = 2 half-lives, so 100 ÷ 4 = 25 counts/s.
Total = 12.5 + 25 = 37.5 counts/s.
💡 Notice that Q started much weaker but is now the bigger part. Long half-lives win in the end.
Next: M1 Units · All ESAT topics