AQA GCSE Combined Science: Trilogy Physics Paper 1 (Higher), 2019: Question 6

💡 Key knowledge

• Gamma radiation is electromagnetic radiation with no mass and no charge.
• When gamma is emitted, the nucleus does not change mass number or charge.
• Gamma is weakly ionising but very penetrating.
• Half-life is the time taken for the number of undecayed nuclei or the activity to halve.

🧠 Exam technique

• Use short, precise science statements for 1-mark answers.
• For calculations, show the pattern clearly: halve, compare, or use gradient.
• If the question says “use the graph”, take values from the graph or a tangent, not from memory.

❌ Common errors

• Saying gamma changes the nucleus mass number or charge.
• Confusing activity with count rate: count rate is often lower because not all radiation is detected.
• Forgetting units, especially Bq and seconds.
• Drawing a line instead of a tangent for a rate from a curve.

✅ Correct answer

Mass number stays the same and charge stays the same.

💡 Key knowledge

• Gamma is just energy leaving the nucleus.
• No protons or neutrons are lost or gained.
• So the atomic number and mass number do not change.

❌ Common errors

• Saying the mass number decreases.
• Saying the charge decreases.
• Mixing gamma with alpha or beta decay.

✅ Correct answer

Gamma radiation is only weakly ionising, so it produces very few ion pairs in a detector.

Alternative acceptable idea: most gamma radiation passes through the detector because it is very penetrating.

💡 Key knowledge

• Ionising radiation causes ion pairs in a detector.
• Weakly ionising means fewer ions are made, so fewer detections happen.
• Gamma can pass through materials more easily than alpha or beta.

🧠 Exam technique

• Use the phrase weakly ionising for full credit.
• A very good extra phrase is very penetrating .

✅ Correct answer

• The radiation spreads out in all directions.
• Only some of the radiation goes into the G-M tube.
• Only some of the radiation entering the G-M tube is detected because gamma is weakly ionising.

💡 Key knowledge

• Activity = number of decays per second from the source.
• Count rate = number of detected counts per second.
• Not every emitted gamma photon is counted.

❌ Common errors

• Saying the source is “less active” because the count rate is lower.
• Talking about contamination instead of detection.
• Forgetting that emissions spread out in all directions.

✅ Correct answer

To reduce the amount of radiation received because radiation increases the risk of cancer or genetic mutation.

💡 Key knowledge

• Radiation dose falls with distance from the source.
• Less exposure means less chance of cell damage.
• Gamma is penetrating, so distance is one useful safety measure.

🧠 Exam technique

• Give both the safety action and the reason for it.
• Good wording: “reduce irradiation” or “reduce exposure”.

📐 Calculations / data check

Step 1: Multiply thickness by count rate.

• 0.5 × 110 = 55
• 1.0 × 60 = 60
• 1.5 × 33 = 49.5
• 2.0 × 18 = 36
• 2.5 × 10 = 25

Step 2: If count rate were inversely proportional to thickness, these products would be constant.

Step 3: They are not the same, so the relationship is not inverse proportion.

✅ Correct answer

The values of thickness × count rate are different, so the data do not show inverse proportion.

Example: 0.5 × 110 = 55, but 1.0 × 60 = 60, so the product is not constant.

❌ Common traps

• Only checking one pair of values.
• Comparing the raw numbers without using the inverse-proportion rule.
• Writing “it decreases” without explaining why that matters mathematically.

✅ Correct answer

¹⁴⁰₅₇La → ⁰₋₁e + ¹⁴⁰₅₈Ce

This is beta decay. The mass number stays the same, and the atomic number increases by 1.

💡 Key knowledge

• In beta decay, a neutron turns into a proton and an electron.
• The emitted particle is an electron, written as ⁰₋₁e .
• Cerium has atomic number 58.

❌ Common errors

• Getting the electron numbers wrong.
• Changing the mass number to 139 or 141.
• Writing alpha decay instead of beta decay.

📐 Step-by-step method

1. Find the half-life from the graph.
The graph shows the half-life is about 50 s.

2. Find the time difference.
350 s − 100 s = 250 s

3. Convert to half-lives.
250 ÷ 50 = 5 half-lives

4. Halve 5 times.
Ratio = (1/2)⁵ = 1/32

✅ Correct answer

Ratio = 1 : 32

Equivalent answers: 1/32 , 32:1 if clearly explained in context, or 0.031 .

❌ Common errors

• Using 100 s and 350 s as the number of atoms instead of times.
• Forgetting to divide the time difference by the half-life.
• Halving only once or twice instead of 5 times.
• Writing the ratio the wrong way round.

📐 Step-by-step method

1. Draw a tangent to the curve at 20 s.
The tangent should just touch the curve at that point, not cross it.

2. Choose two clear points on the tangent.
Read values from the tangent, not from the curve.

3. Find the gradient.
gradient = Δ number of atoms ÷ Δ time

4. Convert to activity.
The gradient is the activity: about 5.3 × 10²¹ Bq.

✅ Correct answer

Activity = 5.3 × 10²¹ Bq

Acceptable range: about 4.7 × 10²¹ Bq to 5.9 × 10²¹ Bq.

🧠 Exam technique

• Always use a tangent for a curved graph at a point in time.
• Make sure the units are correct: activity is in Bq.
• The graph’s y-axis is in ×10²³ , so watch powers of ten carefully.

❌ Common errors

• Drawing a straight line through the graph instead of a tangent.
• Using points from the curve rather than the tangent.
• Forgetting the power of ten on the y-axis.
• Giving the answer in the wrong units or without units.

💡 What top answers did well

• Used the exact physics idea the mark scheme wanted.
• Kept explanations short and focused.
• For graph questions, used the graph properly rather than guessing.

🧠 How to secure marks quickly

• State the key fact first.
• Add the reason if there are 2 marks.
• For calculations, show each step and final units.

❌ Final reminders

• Gamma: no mass change, no charge change.
• Count rate is not the same as activity.
• Inverse proportion means thickness × count rate is constant.
• Activity from a graph comes from the gradient.