The 3- to 6-mark questions in this topic come in four types, and each type gets marked against the same points every time. Choose one, learn its chain, then fill in the details from the question. Every question here is from a real Edexcel paper.
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The notes: all four chains, how to mark up a question and every worked answer. Or just the questions, with the chains at the top and the answers at the back.
“Use the binding energy per nucleon curve to explain…”
…or why fission or fusion releases energy, or why fusion produces large amounts of energy. Any question with the binding energy per nucleon graph in it is this type. If it says “refer to the graph” or “use values from the graph”, a mark is waiting for numbers read off it.
The one idea behind every answer: energy is released whenever the binding energy per nucleon increases. Fusion moves small nuclei up the left side of the curve; fission moves a large nucleus up the right side. Both head towards iron-56 at the peak, the most stable nucleus.
Fusion
Fission
What happens
Two small mass nuclei join to form a larger nucleus
A large mass nucleus splits into two smaller nuclei (plus 2 or 3 neutrons)
Nucleon number
Increases
Decreases
On the curve
Left side, moving right and up
Right side, moving left and up
Binding energy per nucleon
Increases in both, so total mass decreases and energy is released (ΔE = c²Δm)
2. The binding energy chain: six links
THE BINDING ENERGY CHAIN up the curve → energy out
Say which way each process moves, then why that releases energy
Say it until it sticks: small join · large split · towards iron · BE per nucleon up · mass down · ΔE = c²Δm
1
Fusion: small nuclei joinNucleon number increases, so the nuclei move up the left side of the curve.
2
Fission: a large nucleus splitsNucleon number decreases, so the products move up the right side of the curve.
3
Both move towards iron-56Iron-56 is at the peak: the highest binding energy per nucleon, the most stable nucleus.
4
Binding energy per nucleon increasesThe products are more tightly bound than what you started with.
5
So the mass decreasesThe total mass of the products is less than the mass before (the mass deficit increases).
6
Energy released: ΔE = c²ΔmThe lost mass is released as energy, mostly kinetic energy of the products. Quote values off the graph when asked.
The trap: “binding energy is released”. Binding energy is the energy needed to pull a nucleus apart, not energy stored inside it. Energy is released because the binding energy per nucleon goes up. (Check the definition.)
Fusion only, or fission only? Use links 1 or 2 for the process you’re asked about, then 4, 5 and 6. For “both”, you need 1 and 2.
3. Before you write: mark up the question
Ring · SWhat you start with
Small nuclei (fusion) or one large nucleus (fission). Find them on the curve.
Ring · EWhat you end with
Where the products sit on the curve. Higher up = more binding energy per nucleon.
UnderlineThe trigger
“binding energy per nucleon”, “release energy”, “refer to the graph” or “use values from the graph”.
4. Worked example: fusion and fission
How fusion and fission both release energyEdexcel A Level Physics past paper · 3 marks
Nuclear fusion involves small nucleiS joining to make larger nucleiE. Nuclear fission involves large nucleiS splitting to become smaller nucleiE. Both of these processes release energy.
(a) The graph shows how the binding energy per nucleon varies with nucleon number for a range of isotopes.
Both processes move up the curve, towards iron-56 at the peak.
Use the binding energy per nucleon curve to explain how fusion and fission both release energy. (3)
The trigger“binding energy per nucleon curve” and “release energy”. Run the binding energy chain.
Start and endS = small nuclei for fusion, a large nucleus for fission. E = where the products land: higher on the curve both times.
Three marks
“both” = links 1 and 2
then links 4–6 for the energy
no numbers needed: the curve has no scale
Your turn Have a go before you look
Your answer stays on this device, so you can come back to it.
Step 1 Run the chain in note form
1fusion: small join → nucleon no. up → BE/nucleon up2fission: large splits → nucleon no. down → BE/nucleon up3both towards iron-564–6BE/nucleon up → mass down → energy released, ΔE = c²Δm
Step 2 Write it in full sentences
1In fusion, small nuclei join, so the nucleon number increases. On the left of the curve this means the binding energy per nucleon increases.✓ 1 mark
2In fission, a large nucleus splits, so the nucleon number decreases. On the right of the curve this also means the binding energy per nucleon increases.✓ 1 mark
3Both processes move towards iron-56, at the peak of the curve.shows you understand
4When the binding energy per nucleon increases, the products have less mass than the starting nuclei, so energy is released (ΔE = c²Δm).✓ 1 mark
Three marks: one for each direction on the curve, one for why that releases energy.
Check it against the mark scheme
Mark scheme says
Link
Fusion involves an increase in binding energy (per nucleon) as the number of nucleons increases
1
(1)
Fission involves an increase in binding energy (per nucleon) as the number of nucleons decreases
2
(1)
If binding energy per nucleon increases, energy is released in the process
4–6
(1)
3 marks. “Larger nuclei” and “smaller nuclei” are accepted for the nucleon number increasing and decreasing.
5. Your turn: practise the chain
Mark up the question first (ring what you start with and what you end with, underline the trigger), write your answer, then open the answer to see the mark-up, the chain and the mark scheme.
Practice 1 · Why fusion produces so much energyEdexcel A Level Physics past paper · 3 marks
A fusion research centre was opened in Rotherham in 2021. The centre has a device which tests materials in the extreme conditions found inside a fusion reactor.
(b) The graph shows how the binding energy per nucleon varies with nucleon number for a range of nuclides.
Read two values off the graph and compare them: that’s the “refer to the graph” mark.
Explain why the fusion of nuclei can produce large amounts of energy. Your answer should refer to information from the graph. (3)
The trigger“fusion” and “refer to information from the graph”. Binding energy chain, fusion half only.
Start and endS = small nuclei such as hydrogen (left of the curve). E = a larger nucleus such as helium, much higher up.
Three marks
link 1, then links 4–6
one mark is for values from the graph
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
1small nuclei fuse → larger nucleus4–6BE/nucleon increases → mass decreases → energy released6graph: ~1 MeV (²H) → ~7 MeV (⁴He): a big jump
Step 2 Write it in full sentences
1Small (light) nuclei fuse to form a larger nucleus.✓ 1 mark
2The binding energy per nucleon increases, so the mass decreases and energy is released (ΔE = c²Δm).✓ 1 mark
3From the graph, the binding energy per nucleon rises from about 1 MeV for hydrogen-2 to about 7 MeV for helium-4: an increase of about 6 MeV for every nucleon. The curve is steepest here, so fusing light nuclei releases a lot of energy per nucleon.✓ 1 mark
“Refer to the graph” means numbers. Any two sensible values read off the curve, compared, earn the mark.
Check it against the mark scheme
Mark scheme says
Link
Small(er) nuclei fuse to form larger nuclei
1
(1)
The binding energy (per nucleon) increases, releasing energy, or the mass deficit increases and E = c²Δm
4–6
(1)
Comparison of values from the graph, e.g. H to He is about 7 MeV
6
(1)
3 marks. The mark scheme’s own example compares hydrogen and helium. Any correct pair of values from the curve works.
Practice 2 · Fission of uranium-238Edexcel International A Level past paper · 5 marks
The graph shows how the binding energy per nucleon varies with number of nucleons.
Read both values across to the axis: about 7.5 MeV for uranium-238 and 8.5 MeV for tin-120.
(a) Explain why energy is released when a nucleus undergoes fission. (2)
(b) A nucleus of ²³⁸US undergoes fission. One of the fission products is SnE.
Estimate the energy, in MeV, released in the fission. You should use values from the graph. (3)
The trigger“fission” and “use values from the graph”. Binding energy chain, fission half, then a number.
Start and endS = uranium-238, at 7.5 MeV per nucleon. E = tin, at 8.5 MeV per nucleon: higher up.
Five marks
(a) link 2, then link 4
(b) the increase per nucleon × 238 nucleons
it says estimate: one or two sig figs is fine
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
2(a) large nucleus splits4(a) BE/nucleon increases → energy released6(b) 7.5 → 8.5 MeV: up 1.0 MeV per nucleon6(b) × 238 nucleons ≈ 240 MeV
Step 2 Write it in full sentences
(a) Why energy is released
2In fission, a large nucleus splits into two smaller nuclei (and some neutrons).✓ 1 mark
4The binding energy per nucleon increases, so the total binding energy increases and the mass decreases: energy is released.✓ 1 mark
(b) The estimate
6From the graph: binding energy per nucleon of ²³⁸U ≈ 7.5 MeV and of Sn ≈ 8.5 MeV.✓ 1 mark
6Increase in binding energy per nucleon = 8.5 − 7.5 = 1.0 MeV, for each of the 238 nucleons: increase in binding energy ≈ 238 × 1.0 MeV.✓ 1 mark
6Energy released ≈ 240 MeV✓ 1 mark
Assume every nucleon ends up about as tightly bound as tin. That’s why it’s an estimate.
Check it against the mark scheme
Mark scheme says
Link
(a) A large nucleus splits (into two nuclei plus neutrons)
2
(1)
(a) The binding energy (per nucleon) increases, so energy is released, or there is a decrease in (total) mass
4
(1)
(b) Binding energy per nucleon read from the graph for before and after fission
6
(1)
(b) Binding energy before and after fission calculated, or the binding energy increase calculated
“Explain the conditions required to bring about and maintain fusion…”
…in a star, in the Sun, or in a reactor on Earth (JET, ITER, a hot plasma). It’s often the starred 6-marker. Shorter versions ask why each condition is needed, or just to describe the extreme conditions.
Two conditions, two jobs. A very high temperature brings fusion about: the nuclei move fast enough to get close despite repelling each other. A very high densitymaintains it: collisions happen more frequently, so enough fusions keep happening. “Frequently” is the key word. (See the definition.)
Condition
Why it’s needed
Which part of the question
Very high temperature (about 10⁷ K)
Nuclei are positive and repel. They need a high kinetic energy to get close enough (about 10⁻¹⁵ m) for the strong force to pull them together
“bring about”
Very high density (or pressure)
Lots of nuclei close together, so collisions happen more frequently and the fusion rate is high enough to keep going
“maintain”
2. The fusion conditions chain: six links, six marks
THE FUSION CHAIN hot → close → often
Temperature brings it about; density keeps it going
Say it until it sticks: very hot · positive nuclei repel · high KE overcomes repulsion · close enough to fuse · very dense · collisions more frequent
1
Very high temperatureAbout 10⁷ K or more: the material is a plasma of nuclei and electrons.
2
Nuclei are positive, so they repelThere is a large electrostatic repulsive force between the nuclei.
3
High kinetic energy overcomes the repulsionAt a high temperature the nuclei move fast enough to overcome the repulsion.
4
Close enough to fuseThey get within about 10⁻¹⁵ m, where the strong force acts and they fuse.
5
Very high densityLots of nuclei in a small volume (high pressure in a star).
6
Collisions happen more frequentlyThe collision rate is high enough to sustain fusion, and the energy released keeps the temperature high.
The trap: “the nuclei need lots of energy to fuse”. That misses three marks. Say why: they’re positive, so they repel; the energy is kinetic energy from the high temperature; and they must get close enough for the strong force.
No spare link. The starred mark scheme has exactly six points. Six points earn 4 marks, and the last 2 come from writing them as a linked, logical answer, which is exactly what the chain is.
3. Before you write: mark up the question
Ring · WWhere it happens
A star’s core, the Sun, a reactor on Earth. A reactor adds magnetic fields and a plasma.
Ring · HHow many conditions
“the conditions” = both. “each of these two” = both, with a reason for each.
UnderlineThe command word
“bring about and maintain”, “explain why” (give reasons) or “describe” (just say what they are).
4. Worked example: fusion in stars
The conditions for fusion in starsEdexcel A Level Physics past paper · 6 marks
Two positive nuclei repel. Only fast-moving nuclei get close enough for the strong force to take over.
The energy radiated by starsW is released by nuclear fusion.
* Explain the conditionsH required to bring about and maintain nuclear fusion in stars. (6)
The trigger“conditions” and “bring about and maintain fusion”. Run the fusion chain.
Where and how manyW = a star’s core. H = “the conditions”: both temperature and density.
Six marks
the * means your writing is marked too
links 1–4: bring about (temperature)
links 5–6: maintain (density)
Your turn Have a go before you look
Your answer stays on this device, so you can come back to it.
Step 1 Run the chain in note form
1very high temperature2nuclei positive → electrostatic repulsion3high KE → overcome repulsion4close enough → strong force → fuse5very high density6collisions more frequent → sustained
Step 2 Write it in full sentences
Bringing fusion about
1Fusion requires a very high temperature (about 10⁷ K).✓ 1 mark
2All nuclei are positively charged, so there is a large electrostatic repulsive force between them.✓ 1 mark
3At a very high temperature the nuclei have a high kinetic energy, enough to overcome the repulsion.✓ 1 mark
4So the nuclei get close enough (about 10⁻¹⁵ m) for the strong force to act, and they fuse.✓ 1 mark
Maintaining fusion
5Fusion also requires a very high density.✓ 1 mark
6With so many nuclei close together, collisions happen more frequently, so the collision rate is high enough to sustain fusion.✓ 1 mark
Six points in a logical order: 4 marks for the points and 2 for the reasoning. Leaving out any one point costs you a mark.
Check it against the mark scheme
Mark scheme says
Link
Requires a (very) high temperature
1
(1)
Nuclei all have positive charge, leading to a large repulsive force between nuclei
2
(1)
At high temperature nuclei have high kinetic energy, sufficient to overcome repulsion
3
(1)
Nuclei must get close enough to fuse (accept close enough for the strong force)
4
(1)
Requires (very) high density
5
(1)
Collision rate must be high enough to sustain fusion
6
(1)
6 marks, indicative content (the * on the question). Six points seen earn 4 marks; the last 2 are for a logical, linked answer. The chain gives you both: every link is a point, and you write them in order.
5. Your turn: practise the chain
Same chain, different set-ups. Mark up the question first, write your answer, then open the answer to see the mark-up, the chain and the mark scheme.
Practice 1 · The ITER plasmaEdexcel International A Level past paper · 4 marks
(b) It is hoped that the ITER will be the first fusion device on Earth to maintain nuclear fusion for long periods of time.
To achieve fusion in the ITERW, a hot plasma is used. The plasma must
• be at an extremely high temperature • have sufficient density for fusion.
Explain why each of these two conditionsH must be met. (4)
The trigger“explain why each of these two conditions”. Fusion chain: a reason for each.
Where and how manyW = a reactor on Earth, but the physics is the same as a star. H = both conditions, with a reason for each.
Four marks
the conditions are given: don’t just repeat them
temperature: links 2–4
density: link 6
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
3high temp → nuclei have enough KE2…to overcome electrostatic repulsion4…so they get close enough to fuse6density → collision rate high enough6…which maintains the temperature
Step 2 Write it in full sentences
Extremely high temperature
3The temperature must be extremely high so that the nuclei have enough kinetic energy…✓ 1 mark
2…to overcome the electrostatic repulsion between the positively charged nuclei…✓ 1 mark
4…so that the nuclei come close enough to fuse.✓ 1 mark
Sufficient density
6The density must be high enough that collisions happen frequently: the collision rate is high enough for fusion to continue, and the energy released keeps the temperature very high.✓ 1 mark
Five marking points for four marks, so you have one spare. Don’t just repeat the two bullet points: the marks are all in the reasons.
Check it against the mark scheme
Mark scheme says
Link
Very high temperature so that the nuclei have sufficient kinetic energy
3
(1)
Nuclei must overcome electrostatic repulsion (allow repulsion due to positively charged nuclei)
2
(1)
So that the nuclei come close enough to fuse
4
(1)
Sufficient density so that the collision rate (between nuclei) is high (enough)
6
(1)
Sufficient collision rate to maintain the (very high) temperature
6
(1)
Max 4 marks from these 5 points.
Practice 2 · Inside a fusion reactorEdexcel A Level Physics past paper · 3 marks
A fusion research centre was opened in Rotherham in 2021. The centre has a device which tests materials in the extreme conditions found inside a fusion reactorW.
(a) Describe the extreme conditionsH inside a fusion reactor. (3)
The trigger“describe the extreme conditions”. Fusion chain, links 1 and 5, plus what a reactor adds.
Where and how manyW = a reactor, not a star. H = three marks: name three conditions.
Describe, not explain
just say what the conditions are
a reactor adds magnetic fields, neutrons and a plasma
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
1very high temperature, ~10⁷–10⁸ Kplasma: fully ionisedvery strong magnetic fieldbombarded by fast neutrons
Step 2 Write it in full sentences
1A very high temperature, about 10⁷–10⁸ K.✓ 1 mark
+The material is a plasma: it is fully ionised.✓ 1 mark
+There is a very high magnetic flux density (strong magnetic field), which holds the plasma away from the walls.✓ 1 mark
+Materials are bombarded by neutrons released in the fusion reactions.a fourth point, spare
“Describe” is a list of what’s there. Very high density or pressure also scores, but the reactor-only points are safer.
Check it against the mark scheme
Mark scheme says
Link
Very high temperatures (accept T ≈ 10⁷–10⁸ K)
1
(1)
High magnetic flux density
+
(1)
Bombardment by neutrons
+
(1)
Material is a plasma, or material is fully ionised
+
(1)
Max 3 marks. Conditions for stars are accepted too: high density, high pressure.
Learn the definition word for word.“Very high density, so collisions happen more frequently”: the conditions for fusion and the other Topic 11 definitions, with a wall poster.
“Explain why there is an exponential decrease in the rate of decay…”
…or how a decay constant can be found when decay is random, or how a radioactive sample keeps releasing energy for a long time. The clues: random, a large number of nuclei, exponential, half-life, rate of decay.
Random for one nucleus, predictable for billions. You can’t predict which nucleus will decay next or when. But every nucleus has the same fixed chance of decaying each second: the decay constant λ. With a huge number of nuclei, the fraction that decays each second is λ, so the pattern is predictable. (Random and spontaneous: the definitions.)
2. The decay chain: six links, six marks
THE DECAY CHAIN chance → pattern → exponential
Start with one nucleus, end with the whole sample
Say it until it sticks: fixed probability · large number → predictable · fraction = λ · A = λN · N falls · A falls exponentially
1
Fixed probability (λ)Each nucleus has a fixed probability of decaying per second: the decay constant λ.
2
Large number → predictableDecay is random for one nucleus, but a large number of nuclei gives a predictable pattern.
3
Fraction decaying per second = λSo the same fraction of the remaining nuclei decays every second.
4
Activity depends on N: A = λNThe number decaying per second depends on the number of unstable nuclei left.
5
N decreases exponentiallyAs nuclei decay there are fewer left: N = N₀e−λt.
6
So the rate of decay decreases exponentiallyA = A₀e−λt: it halves every half-life.
The trap: “the nuclei decay more slowly as they get older”. They don’t. The probability λ stays the same for every nucleus. The rate falls only because there are fewer nuclei left to decay.
3. Before you write: mark up the question
Ring · LHow many nuclei
“a large number of unstable nuclei”: that’s your permission to use probability.
Ring · QWhat you’re explaining
The rate (activity), the number of nuclei, the decay constant, or the energy released.
UnderlineThe trigger
“exponential”, “random”, “decay constant”, “half-life”, “over a long period of time”.
4. Worked example: the exponential decrease
Why the rate of decay falls exponentiallyEdexcel A Level Physics past paper · 6 marks
Radioactive decay is often described in textbooks as a spontaneous, random process.
The same fraction decays each second, so the activity halves every half-life.
* (b) Explain why there is an exponential decrease in the rate of decayQ for a sample containing a large number of unstable nucleiL. (6)
The trigger“exponential decrease” and “random”. Run the decay chain.
Number and quantityL = a large number of nuclei: so the pattern is predictable. Q = the rate of decay (activity).
Six marks
the * means your writing is marked too
six links, six points: none spare
equations count as points: A = λN, N = N₀e−λt
Your turn Have a go before you look
Your answer stays on this device, so you can come back to it.
Step 1 Run the chain in note form
1fixed probability per second (λ)2large number → predictable pattern3fraction decaying per second = λ4A = λN5N = N₀e−λt6A = A₀e−λt
Step 2 Write it in full sentences
1Each nucleus has a fixed probability of decaying in the next second: the decay constant λ.✓ 1 mark
2Decay is random for an individual nucleus, but for a sample with a large number of unstable nuclei there is a predictable pattern.✓ 1 mark
3The fraction of the nuclei decaying in the next second is equal to the decay constant λ.✓ 1 mark
4So the number of nuclei decaying per second depends on the number of unstable nuclei left: A = λN.✓ 1 mark
5As nuclei decay, the number of unstable nuclei decreases exponentially with time: N = N₀e−λt.✓ 1 mark
6So the rate of decay decreases exponentially with time: A = A₀e−λt.✓ 1 mark
Each link leads into the next. That chain of reasoning is what earns the 2 marks for structure.
Check it against the mark scheme
Mark scheme says
Link
There is a fixed probability (λ) of an individual nucleus undergoing decay (in the next second)
1
(1)
For a sample with a large number of unstable nuclei there is a predictable pattern
2
(1)
The fraction of nuclei decaying in the next second is equal to the decay constant (λ)
3
(1)
Hence the number of nuclei decaying (in the next second) depends on the number of (unstable) nuclei, or activity = λN
4
(1)
The number of unstable nuclei decreases exponentially (with time), or N = N₀e−λt
5
(1)
So the rate of decay decreases exponentially (with time), or A = A₀e−λt
6
(1)
6 marks, indicative content (the * on the question). Six points seen earn 4 marks; the last 2 are for a logical, linked answer. The chain gives you both: every link is a point, and you write them in order.
5. Your turn: practise the chain
Same chain, used in two different ways. Mark up the question first, write your answer, then open the answer to see the mark-up, the chain and the mark scheme.
Practice 1 · Finding the decay constantEdexcel A Level Physics past paper · 2 marks
Nuclear decay is described as being spontaneous and random.
(a) (ii) Explain why the decay constantQ of an isotope can be determined even though nuclear decay is random. (2)
The trigger“decay constant” and “random”. Decay chain, links 1 and 2.
Number and quantityQ = the decay constant. L isn’t in the question, so you add it: a very large number of nuclei.
Two marks
two marks = links 1 and 2
the first two links are always the same
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
1λ = probability of decay per second: can be found2very large numbers of nuclei2 marks: stop here
Step 2 Write it in full sentences
1The decay constant is the probability that a nucleus decays per unit time, and this probability can be determined…✓ 1 mark
2…because a sample contains a very large number of nuclei, so the fraction decaying per second is predictable even though each decay is random.✓ 1 mark
Two marks, two links. The idea is in the first two links of every decay answer.
Check it against the mark scheme
Mark scheme says
Link
The probability of decay (in a specified time) may be determined
1
(1)
(The probability may be applied accurately because) very large numbers are involved
2
(1)
2 marks.
Practice 2 · Warm radiumEdexcel A Level Physics past paper · 6 marks
* Radium is a radioactive element. The most common isotope of radium has a half-life of almost two thousand years. A sample of radium can remain at a higher temperature than its surroundings for a long period of time.
Explain how a sample of radium is able to release significant amounts of energyQover a long period of time. (6)
The trigger“over a long period of time” and “half-life”. Decay chain link 4, plus where the energy comes from.
Number and quantityQ = energy, so you need ΔE = c²Δm. The long half-life explains “a long period of time”.
Six marks
the * means your writing is marked too
two halves: why it lasts and where the energy comes from
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
4A = λN: rate depends on Nlong half-life → nuclei there for a long timeeach decay: small decrease in massΔE = c²ΔmΔm small, c² huge → lots of energyenergy → KE of atoms → sample stays warm
Step 2 Write it in full sentences
Why it lasts a long time
4The rate of decay depends on the number of unstable nuclei in the sample: A = λN.✓ 1 mark
+Radium has a long half-life (small λ), so unstable nuclei are present for a long time and decays keep happening.✓ 1 mark
Where the energy comes from
+When a nucleus decays, there is a small decrease in mass.✓ 1 mark
+Energy is released according to ΔE = c²Δm.✓ 1 mark
+Δm is small but c² is very large, so a significant amount of energy is released.✓ 1 mark
+The energy released becomes kinetic energy of the atoms in the sample, so the sample stays above the temperature of its surroundings.✓ 1 mark
Only one link comes straight from the chain. The other five are about energy: mass deficit and ΔE = c²Δm.
Check it against the mark scheme
Mark scheme says
Link
The rate of decay depends upon the number of unstable nuclei in the sample (A = λN)
4
(1)
Radium has a large half-life, so unstable nuclei are present in the sample for a long time
+
(1)
When a nucleus decays there is a (small) decrease in mass
+
(1)
Energy is released according to ΔE = c²Δm
+
(1)
Δm is small but c is large, so a significant amount of energy is released
+
(1)
Energy released by the decay becomes kinetic energy of the atoms in the sample (hence the sample is above the temperature of the surroundings)
+
(1)
6 marks, indicative content (the * on the question). Six points seen earn 4 marks; the last 2 are for a logical, linked answer. The chain gives you both: every link is a point, and you write them in order. A reference to binding energy increasing is accepted for the decrease in mass.
Learn the words the mark scheme uses.Random, spontaneous, activity and half-life, word for word, with a wall poster.
…or describe how to find the type, or the fraction, of radiation from a source using a Geiger–Müller tube. The clues: GM tube, ratemeter, count rate, paper, aluminium, lead, background.
Radiation
Stopped by
Why
Alpha (α)
A sheet of paper, or a few cm of air
Most strongly ionising, so it loses its energy quickly
Beta (β)
A few mm of aluminium
Less ionising than alpha
Gamma (γ)
Never fully stopped: several cm of lead reduces it
Weakly ionising
Background first, every time. The GM tube counts background radiation too. Measure it with the source away, and take it off every reading before you compare.
2. The absorber chain: six links
THE ABSORBER CHAIN background → absorber → compare
Measure, put each absorber in, then reason from what gets through
Say it until it sticks: background · no absorber · paper stops α · aluminium stops β · γ gets through · compare corrected counts
1
Measure the background count rateWith the source removed. Take it off every reading (corrected count rate).
2
Measure the count rate with no absorberSource and GM tube at a fixed distance, nothing in between.
3
Paper stops alphaIf the count rate doesn’t fall with paper in place, there is no alpha.
4
A few mm of aluminium stops betaIf the count rate falls to background with aluminium, there is beta and no gamma.
5
Gamma passes through bothIf counts above background are left after aluminium, there is gamma.
6
Compare the corrected count ratesSay what each change shows, or work out a fraction: (count with absorber − background) ÷ (count without − background).
The trap: only saying what it IS. “It’s beta because aluminium stopped it” scores one mark. Rule out the other two as well, each with its evidence: that’s where the rest of the marks are.
3. Before you write: mark up the question
Ring · AEach absorber
Paper, the thickness of aluminium, lead. One link for each.
Ring · CWhat the count did
“no change”, “decreased to background”, “decreased but stayed above background”.
UnderlineThe command word
“Explain which” (reason from results) or “Describe how” (write the method).
4. Worked example: the unknown source
Which radiation is it?Edexcel International A Level past paper · 4 marks
The penetration of nuclear radiation through materials depends upon the ionising power of the radiation.
(a) A teacher tested an unknown radioactive source. He set up the source and a GM tube, as shown.
Each absorber goes between the source and the GM tube.
The initial count rate is determined without an absorber between the GM tube and the source.
When an aluminium sheet of thickness 0.5 cmA was placed between the source and the GM tube, the count rate decreased to background levelC.
The teacher then removed the aluminium sheet and replaced it with a sheet of paperA. There was no changeC in the initial count rate.
Explain which type of radiation is emitted from the source. You should refer to the penetration of each type of radiation through each sheet of absorber. (4)
The trigger“explain which type of radiation”. Absorber chain, links 3–5.
Absorbers and countsA = aluminium, then paper. C = down to background with aluminium; no change with paper.
Four marks
rule out alpha: 2 marks
rule out gamma: 1 mark
name beta, with its reason: 1 mark
Your turn Have a go before you look
Your answer stays on this device, so you can come back to it.
Step 1 Run the chain in note form
3α stopped by paper3no change with paper → no α5γ would pass through Al → count fell to background → no γ4β stopped by Al → it’s β
Step 2 Write it in full sentences
3Alpha radiation would not pass through the sheet of paper.✓ 1 mark
3There was no change in the count rate when the paper was placed between the source and the GM tube, so there can’t be any alpha radiation.✓ 1 mark
5Gamma radiation would pass through the aluminium sheet, but the count rate fell to background, so there can’t be any gamma radiation.✓ 1 mark
4So it must be beta radiation: beta passes through paper but not through 0.5 cm of aluminium.✓ 1 mark
One sentence per absorber, then the conclusion: the question asked you to refer to each sheet.
Check it against the mark scheme
Mark scheme says
Link
Alpha radiation would not pass through the sheet of paper
3
(1)
No change in count rate when paper placed (between source and GM tube), so there can’t be any alpha radiation
3
(1)
There can’t be any gamma radiation, as gamma radiation would pass through the aluminium sheet, or as the count rate decreases to background with the aluminium sheet
5
(1)
It must be beta radiation, as beta radiation would not pass through the aluminium sheet
4
(1)
4 marks.
5. Your turn: practise the chain
This time you write the method. Mark up the question first, write your answer, then open the answer to see the mark-up, the chain and the mark scheme.
Practice 1 · How much of it is beta?Edexcel A Level Physics past paper · 4 marks
(b) A radioactive source used in a school laboratory emits alpha and beta radiationA.
Paper takes out the alpha. Whatever is left above background is beta.
Describe howthe percentage of the activity due to beta radiationC may be determined using a Geiger–Müller tube and ratemeter. (4)
The trigger“describe how”, with a GM tube. Absorber chain, written as a method.
Absorbers and countsA = alpha and beta only, so you need paper. C = a percentage: a ratio of two corrected count rates.
Four marks
links 1, 2 and 3
then the percentage: link 6
no gamma here, so no lead needed
Your turn Write your answer
Your answer stays on this device.
Step 1 Run the chain in note form
1background count rate2count rate from the source, no absorber3paper in: absorbs α6% = (with paper − bg) ÷ (without − bg) × 100
Step 2 Write it in full sentences
1Measure the background count rate with the source removed.✓ 1 mark
2Measure the count rate from the source, with nothing between it and the GM tube.✓ 1 mark
3Place a sheet of paper between the source and the GM tube to absorb the alpha radiation, and measure the count rate again.✓ 1 mark
6Percentage due to beta = (count rate with paper − background) ÷ (count rate without paper − background) × 100.✓ 1 mark
Keep the distance between the source and the GM tube the same for every reading, so only the absorber changes.
Check it against the mark scheme
Mark scheme says
Link
Measure the background count (rate)
1
(1)
Measure count (rate) from source
2
(1)
Use paper to absorb alpha radiation
3
(1)
Percentage = (count with paper − background rate) ÷ (count from source − background rate) × 100
6
(1)
4 marks.
Test yourself on the whole topic.Topic 11 as a self-marking test: decay, half-life, binding energy, fission and fusion.
