What are the key components of Practical Skills II - Implementation and Measurement in Edexcel International A Levels Physics?

Practical Skills II - Implementation and Measurement focuses on the ability to accurately implement experimental procedures and measure physical quantities. Key components include understanding experimental design, selecting appropriate equipment, ensuring precise and accurate measurements, and analyzing data effectively.

How can I improve my measurement accuracy in Practical Skills II for Edexcel Physics?

To improve measurement accuracy, ensure that you use calibrated instruments, take multiple readings to calculate an average, minimize parallax errors by reading measurements at eye level, and use appropriate significant figures. Understanding the limitations of your measuring instruments and applying error analysis techniques are also crucial.

Why is error analysis important in Practical Skills II - Implementation and Measurement?

Error analysis is essential because it helps identify the reliability and precision of your measurements. By understanding and calculating errors, you can assess the quality of your data, make necessary adjustments, and provide a more accurate interpretation of experimental results. This is a critical skill in Edexcel International A Levels Physics.

What types of equipment are commonly used in Practical Skills II - Implementation and Measurement?

Common equipment includes rulers, calipers, micrometers for length measurements; stopwatches and timers for time; voltmeters and ammeters for electrical measurements; and thermometers for temperature. Each piece of equipment requires understanding its specific use and limitations to ensure accurate data collection.

How does Practical Skills II - Implementation and Measurement prepare students for further studies in Physics?

This subtopic equips students with essential skills for conducting experiments, analyzing data, and understanding the scientific method. These skills are foundational for advanced studies in Physics and other scientific disciplines, as they foster critical thinking, problem-solving, and the ability to conduct independent research.

Why is "Not subtracting the background count from a measured count rate" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students record the count with the source present and quote it directly, forgetting that background radiation is always being detected too. How to avoid it: Always measure the **background count rate with no source present** and subtract it: **corrected rate = measured rate − background rate**. Omitting it is a systematic error that inflates the source activity. Source: Edexcel IAL Physics Unit 6 examiner reports

Why is "Forgetting to multiply the number of divisions by the time/div when reading a time off an oscilloscope" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students read the number of divisions and treat it directly as the time, ignoring the time-base setting. How to avoid it: A time interval is **t = (horizontal divisions) × (time/div)**. Read the time-base setting off the oscilloscope and multiply by it every time, then convert to seconds. Source: Edexcel IAL Physics Unit 6 examiner reports

Why is "Timing a single oscillation instead of many, then dividing" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students start and stop the timer on one swing, so the ~0.2 s reaction error is a large fraction of the period. How to avoid it: Time **many oscillations** (e.g. 20) and divide by the number: **T = t / n**. This spreads the timing error over all the swings (or use light gates to remove reaction time). Source: Edexcel IAL Physics Unit 6 examiner reports

Why is "Recording readings to the wrong (or inconsistent) number of significant figures" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students copy whatever the display shows without matching it to the instrument's resolution, mixing precisions within a table; ln/log columns are rounded inconsistently. How to avoid it: Record every reading from one instrument to a **consistent** number of significant figures set by its resolution (e.g. 2 d.p. in mm for a micrometer), and quote ln/log columns to **3 s.f.** Do not over-state precision the scale cannot support. Source: Edexcel IAL Physics Unit 6 examiner reports

Why is "Ignoring a zero error, or getting its sign wrong (adding when you should subtract)" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students record the raw reading without checking the closed-jaw value, or forget that a negative zero error is added back. How to avoid it: Always check the reading with the instrument set to zero first. Then **corrected = reading − zero error**: a positive zero error is subtracted, a negative one is added back. A zero error is systematic — repeating readings will not remove it. Source: Edexcel IAL Physics Unit 6 examiner reports

Why is "Treating the natural scatter of radioactive counts as a mistake (or as anomalies to be discarded)" a common mistake in Practical Skills II - Implementation and Measurement?

Why it happens: Students expect repeat counts to be identical and label every small difference as an error, not realising decay is a random process. How to avoid it: Remember counts scatter by about **±√N** because decay is random — small differences are **expected statistical variation**, not anomalies. Only discard a reading that lies far outside this spread, and count for longer (larger N) to reduce the relative scatter. Source: Edexcel IAL Physics Unit 6 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Practical Skills II - Implementation and Measurement

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Detailed Study Notes

Detailed notes on Practical Skills in Physics II for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Practical Skills II — Implementation and Measurement (Unit 6, WPH16) — Pearson Edexcel International A Level Physics Study Notes

The IA2 practical-skills toolkit: reading Unit 4/5 instruments to the correct resolution — an oscilloscope or data-logger for capacitor discharge and its time constant, a GM tube and scaler for radioactivity (counts → count rate, subtracting background), a micrometer with a zero-error check, and a digital timer or light gates for the period of oscillation. Recording to consistent significant figures with sensible table headings, distinguishing anomalies from the statistical scatter of counts, separating systematic from random errors, and suggesting specific, named apparatus improvements.

At a glance

This is a skills unit, not new content. WPH16 is a written IA2 exam that tests the same practical skills as Unit 3, but with Unit 4/5 instruments and contexts (capacitor discharge, radioactivity, oscillations).
Count rate = counts ÷ time. Always convert raw counts to a rate and state the unit (s⁻¹ or per minute).
Always subtract the background. Corrected count rate = measured rate − background rate. Skipping this is a systematic overestimate of the source.
Radioactive counts scatter by about ±√N because decay is random. Small differences between repeat counts are expected, not anomalies.
Oscilloscope time = divisions × time/div. The time-base gives times (a period, a discharge time, a time constant); the Y-gain (volts/div) gives p.d.
Time constant τ is the time for the p.d. to fall to 1/e ≈ 0.37 of its start value — read it off the trace as divisions × time/div.
Period from many oscillations: T = t / n. Timing 20 swings and dividing spreads the reaction-time error; light gates remove it.
Check for a zero error on a micrometer/caliper first: corrected = reading − zero error. A systematic error is not reduced by repeating.
Record to consistent significant figures set by the instrument's resolution; log ln/log columns to 3 s.f.; head every table column with quantity / unit.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

6.1 — Read IA2 instruments to their correct resolution: an oscilloscope/data-logger for a capacitor-discharge trace, a GM tube + scaler/ratemeter for radioactivity, a micrometer/caliper for small dimensions, and a digital timer/light gates for oscillation period.
6.2 — Convert raw counts to a count rate and correct it for background radiation; appreciate the statistical (±√N) nature of counts.
6.3 — Read a time interval, a discharge time and a time constant off an oscilloscope trace using the time-base.
6.4 — Record data to a consistent number of significant figures with correct units and sensible table headings, including ln/log columns to 3 s.f.
6.5 — Identify anomalies (distinct from statistical scatter), and assess range, precision and accuracy.
6.6 — Distinguish systematic from random errors at IA2 level (unsubtracted background, zero error, voltmeter loading, reaction-time error) and suggest specific, named apparatus improvements.

Radioactivity: counts, count rate and background

Turn counts into a rate, then subtract the background — and expect ±√N scatter.

In Unit 5 you meet radioactivity; in Unit 6 you must handle its measurements correctly. A Geiger–Müller (GM) tube connected to a scaler (which totals counts) or a ratemeter (which shows a rate directly) detects ionising radiation.

From counts to count rate. A scaler records the total number of counts in a fixed time. The count rate is $\text{count rate} = \frac{\text{counts}}{\text{time}}$ quoted in s⁻¹ (counts per second) or counts per minute. Always divide by the time and state the unit — a bare count is not a rate.

Subtracting the background. Ionising radiation is always present (cosmic rays, rocks, building materials, food). This background is detected even with no source. So you must:

Record the background count rate with no source present.
Subtract it from every measured rate: $R = R_{meas} - R_{bg}$ Omitting this step is a systematic error that inflates the source's apparent activity.

Counts are statistical. Radioactive decay is a random process, so successive counts of the same source are not identical. For a count $N$ the natural spread is about $\pm\sqrt{N}$ . So 512, 498, 505 and 507 counts are all consistent (spread of order $\pm\sqrt{500}\approx\pm22$ ) — these are not anomalies. Counting for longer (larger $N$ ) reduces the relative scatter, since $\sqrt{N}/N = 1/\sqrt{N}$ .

A GM tube feeds total counts to a scaler. Divide the counts by the counting time to get a rate, then subtract the background count rate to get the source-only (corrected) count rate. Counts scatter by about ±√N because decay is random.

Count rate = counts ÷ time; state the unit (s⁻¹ or per minute).
Corrected rate = measured rate − background rate; always measure the background.
Counts scatter by about ±√N (random decay) — small differences are not anomalies.
Count for longer (larger N) to reduce the relative statistical scatter.

See the full worked example for practical skills ii - implementation and measurement →

The oscilloscope: reading a capacitor-discharge trace

Time-base gives times; Y-gain gives p.d.; the 0.37V₀ level gives the time constant.

In Unit 4 a capacitor discharges through a resistor and its p.d. falls exponentially. An oscilloscope (or a data-logger) displays this decaying trace, and you must read it correctly.

Two independent settings.

The time-base (time/div) sets how much time each horizontal division represents (e.g. 10 ms per division). A time interval read from the screen is $t = (\text{horizontal divisions}) \times (\text{time/div})$
The Y-gain (volts/div) sets how much p.d. each vertical division represents (e.g. 2.0 V per division). A p.d. is $V = (\text{vertical divisions}) \times (\text{volts/div})$

Reading the time constant off the trace. The time constant $\tau$ is the time for the p.d. to fall to $\tfrac{1}{e} \approx 0.37$ of its initial value $V_0$ . So:

Read the initial p.d. $V_0$ (vertical divisions × volts/div).
Find the point where the trace has fallen to $0.37V_0$ .
Read the time from the start to that point as divisions × time/div — that time is $\tau$ .

Note the Y-gain is only needed to locate the 0.37V₀ level; the time-base does the work of giving $\tau$ . Always convert the time/div to seconds (1 ms = 10⁻³ s) before quoting.

On a capacitor-discharge trace, the time constant τ is the time for the p.d. to fall to 0.37V₀ (that is V₀/e). Locate the 0.37V₀ level using the Y-gain, then read the horizontal time to that point as divisions × time/div. Convert ms to s before quoting.

Time interval t = horizontal divisions × time/div (the time-base).
P.d. V = vertical divisions × volts/div (the Y-gain).
Time constant τ = time for p.d. to fall to 0.37V₀ (= V₀/e), read off the time-base.
Convert time/div (ms/µs) to seconds before quoting a time or frequency.

See the full worked example for practical skills ii - implementation and measurement →

Micrometers, calipers and timing oscillations

Check for zero error, and time many oscillations to shrink the reaction error.

Micrometer and vernier caliper — dimensions of small objects. These measure the diameter of a spring wire, the diameter of a sphere, or the thickness of a gamma absorber to high resolution (micrometer 0.01 mm; typical vernier 0.1 mm). Read them as main scale + (secondary divisions × secondary resolution).

Always check for a zero error first. Close the jaws (or set the instrument to zero) and note the reading. If it is not zero, that is a zero error — a systematic error that shifts every reading. Correct it with $\text{corrected} = \text{reading} - \text{zero error}$ A positive zero error is subtracted; a negative one is added back. Repeating readings does not remove a zero error.

Read a micrometer as main scale plus thimble (divisions × 0.01 mm). Always check the closed-jaw reading first: a zero error is systematic and is corrected by subtracting it (a negative zero error is therefore added back).

Timing oscillations — beat the reaction error. To find the period of a pendulum or a mass–spring system, do not time a single swing: a human reaction time of about 0.2 s would be a large fraction of the period. Instead:

Time many oscillations (e.g. 20) with a digital timer and divide: $T = t/n$ . The reaction error is spread over all $n$ periods.
Better still, use light gates and a data-logger, which trigger electronically and remove reaction time entirely.
Repeat the timing and take a mean, discarding any anomaly.

The frequency then follows from $f = 1/T$ (with $T$ in seconds).

Read micrometer/caliper as main scale + (secondary divisions × resolution).
Check the zero reading first; corrected = reading − zero error (systematic).
Period from many oscillations: T = t / n reduces the reaction-time error.
Light gates + data-logger remove reaction time entirely; f = 1/T.

See the full worked example for practical skills ii - implementation and measurement →

Recording data: significant figures, units and tables

Consistent sig figs set by resolution, quantity/unit headings, ln/log columns to 3 s.f.

Marks are routinely lost not in the physics but in the recording. Edexcel expects disciplined tables.

Consistent significant figures. All readings from one instrument are known to the same precision, so record them to the same number of significant figures (or decimal places), set by the instrument's resolution:

micrometer (0.01 mm) → 2 d.p. in mm (e.g. 1.26 mm, not 1.3 or 1.260)
digital timer → the display's precision (e.g. 27.60 s)
scaler counts → whole numbers; a derived count rate to a sensible 3 s.f.

Mixing 0.5 A with 0.5203 A in one column falsely implies the readings differ in precision and makes anomalies hard to spot.

Table headings carry the unit. Head every column with the quantity and its unit, e.g. Thickness x / mm, Count rate R / s⁻¹, Time t / s. The cells then contain pure numbers. A heading like "Time (s)" is acceptable; a bare "Time" with units written inside every cell is penalised.

Processed (ln / log) columns. For exponential relationships (capacitor discharge, radioactive decay) you often plot a logarithm to get a straight line. For $V = V_0 e^{-t/RC}$ , taking natural logs gives $\ln V = \ln V_0 - \frac{t}{RC}$ so a graph of $\ln V$ against $t$ is a straight line of gradient $-1/(RC)$ . Record such ln (or log) columns to 3 significant figures, headed with the operation, e.g. ln(V / V) (a logarithm has no unit). Keep the same number of significant figures down the whole column.

A disciplined table: consistent significant figures set by the instrument's resolution, headings that carry the quantity and unit, and a processed ln column quoted to 3 s.f. (a logarithm is dimensionless, so it has no unit).

Record to consistent sig figs/decimals set by the instrument resolution.
Head columns with quantity / unit; cells contain pure numbers.
ln/log columns to 3 s.f.; a logarithm is dimensionless (no unit).
For V = V₀e^(−t/RC): plot ln V vs t → straight line, gradient −1/(RC).

Anomalies, systematic vs random errors, and improvements

Name the error type, match it to a remedy, and name a specific better instrument.

The high-tariff Unit 6 questions ask you to evaluate a procedure. Two ideas do most of the work.

Anomaly vs statistical scatter. An anomaly lies well outside the pattern of the other readings and is usually a mistake — identify and exclude it before averaging. But for counts, remember the natural $\pm\sqrt{N}$ scatter: a spread of a few percent is expected, so do not discard readings just because they differ slightly.

Systematic vs random errors (IA2 examples).

Error	Type	Signature	Remedy
Background not subtracted	Systematic	Every rate too high by the same amount	Measure and subtract background rate
Zero error on a micrometer	Systematic	Every reading shifted the same way	Subtract the zero error / re-zero
Voltmeter loading a capacitor circuit	Systematic	Capacitor discharges too fast every time	Use a high-resistance (digital) voltmeter
Reaction time in timing by eye	Random	Times scatter both ways about a mean	Time many oscillations; use light gates
Scatter of radioactive counts	Random	Counts vary ±√N unpredictably	Count for longer (larger N); take a mean

The crucial exam point: repeating readings reduces random error but never removes a systematic error. A precise-but-biased result (small scatter, wrong mean) is the fingerprint of a systematic error, corrected only by re-zeroing, recalibrating, or subtracting the offset.

Suggesting improvements — be specific. Every improvement must name the instrument or technique AND state the specific benefit (higher resolution, or a named error removed). Compare:

✗ "Measure more carefully" — earns nothing.
✓ "Use a micrometer (0.01 mm) instead of a ruler (1 mm) — 100× finer resolution, so a much smaller percentage uncertainty in the thickness."
✓ "Subtract the background count rate — removes the systematic overestimate of the source."
✓ "Time 20 oscillations with light gates and divide — removes reaction time, reducing the random timing error per period."
✓ "Count for several minutes rather than 10 s — larger N, so the relative statistical scatter (1/√N) is smaller."

Anomaly = well outside the pattern → exclude before averaging; count scatter (±√N) is not an anomaly.
Systematic = same shift every time (background, zero error, voltmeter loading); random = scatters both ways.
Repeating reduces random error only — a systematic error survives repetition.
Improvements must name the instrument/technique AND the specific error or resolution benefit.

See the full worked example for practical skills ii - implementation and measurement →

Quick recap

Count rate = counts ÷ time; corrected count rate = measured rate − background rate (always subtract the background).
Radioactive counts scatter by about ±√N — small differences are statistical, not anomalies; count for longer to reduce relative scatter.
Oscilloscope: time = horizontal divisions × time/div; p.d. = vertical divisions × volts/div; time constant τ is the time to fall to 0.37V₀.
Period from many oscillations: T = t / n; light gates remove reaction time; f = 1/T with T in seconds.
Check for a zero error (systematic): corrected = reading − zero error; repeating never removes it.
Record to consistent significant figures set by resolution; head columns with quantity / unit; ln/log columns to 3 s.f.
Repeating reduces random error only; systematic errors (background, zero error, voltmeter loading) need the offset corrected, and improvements must name a specific instrument/technique and benefit.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Count rate $= \dfrac{\text{counts}}{\text{time}}$ ; corrected rate $R = R_{meas} - R_{bg}$ (subtract the background).
Radioactive counts scatter by about $\pm\sqrt{N}$ — statistical, not anomalies.
Oscilloscope time $= (\text{divisions}) \times (\text{time/div})$ ; p.d. $= (\text{divisions}) \times (\text{volts/div})$ .
Time constant $\tau$ = time for p.d. to fall to $\tfrac{1}{e} \approx 0.37$ of $V_0$ (read off the time-base).
Period from $n$ oscillations: $T = \dfrac{t}{n}$ ; then $f = \dfrac{1}{T}$ (T in seconds).
Zero error is systematic: corrected $=$ reading $-$ zero error (negative zero error is added back).
Systematic error: same shift every time, NOT reduced by repeating. Random error: scatters both ways, reduced by repeats/mean.
For $V = V_0 e^{-t/RC}$ : plot $\ln V$ vs $t$ → straight line, gradient $-\dfrac{1}{RC}$ ; ln columns to 3 s.f.

How it’s examined

Unit 6 (WPH16) is a written IA2 practical-skills paper drawing on the Unit 4 and Unit 5 experiments. It is NOT a content-recall paper: it tests whether you can read instruments, record and process data, and evaluate procedures. Typical formats: a read-off from an oscilloscope trace or a scaler (1–2 marks — divisions × time/div, or counts ÷ time with a background subtraction); a table-completion or 'complete the ln column to 3 s.f.' task (1–2 marks); 'identify the anomaly / explain why the spread is not anomalous' for radioactive counts (2–3 marks, AO3); a graph-linearisation item (plot ln V against t, gradient = −1/(RC)); and an extended 'evaluate the procedure / suggest improvements' question (4–6 marks) where you must classify systematic vs random errors and name specific apparatus improvements with their benefit. ALWAYS quote units, subtract the background, keep consistent significant figures, and — for improvements — name the instrument/technique AND the error it removes. Vague answers ('measure more carefully') score nothing.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 6 (WPH16) sample assessment materials and mark schemes; Edexcel IAL Physics Unit 6 examiner reports. Last reviewed 2026-07-06.

Take this whole topic with you

Step-by-step worked examples — Practical Skills II - Implementation and Measurement

Step-by-step solutions to past-paper-style questions on practical skills ii - implementation and measurement, written exactly the way a tutor would explain them at the board.

Question type:

1Count rate from a scaler over a fixed time (2 marks)
Getting startedDirect calculation• Edexcel IAL Physics Unit 6 style• count rate, GM tube, Unit 6
Question
A Geiger–Müller tube connected to a scaler records 4800 counts from a gamma source in a fixed time of 120 s. Calculate the count rate. (2 marks)
Step-by-step solution
1. Step 1
  The count rate is the number of counts recorded divided by the time over which they were counted.
  $\text{count rate} = \frac{\text{counts}}{\text{time}} = \frac{4800}{120}$
2. Step 2
  Evaluate and quote with the correct unit (counts per second, s⁻¹).
  $\text{count rate} = 40\ \text{s}^{-1}$
Answer
Count rate $= 4800 / 120 = 40\ \text{s}^{-1}$ (40 counts per second).
Examiner tip
Marks: (1) counts ÷ time set up; (2) 40 s⁻¹ with the unit. Count rate is a RATE — always divide the raw counts by the counting time, and state the unit (s⁻¹ or counts per minute). Quoting a bare '4800' as the count rate is the classic slip.
2Reading a time interval off an oscilloscope trace (2 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 6 style• oscilloscope, capacitor discharge, time-base, Unit 6
Question
On an oscilloscope displaying a discharging capacitor's p.d., the trace falls to a chosen level over 3.5 horizontal divisions. The time-base is set to 20 ms per division. State the time interval this represents, in seconds. (2 marks)
Step-by-step solution
1. Step 1
  A horizontal interval on the screen is the number of divisions multiplied by the time-base setting (time per division).
  $t = (\text{divisions}) \times (\text{time/div}) = 3.5 \times 20\ \text{ms}$
2. Step 2
  Evaluate, then convert milliseconds to seconds (1 ms = 10⁻³ s).
  $t = 70\ \text{ms} = 70 \times 10^{-3} = 0.070\ \text{s}$
Answer
$t = 3.5 \times 20 = 70\ \text{ms} = 0.070\ \text{s}$ .
Examiner tip
Marks: (1) t = divisions × time/div = 70 ms; (2) converted to 0.070 s. Forgetting to multiply by the time/div (quoting t = 3.5 s) and leaving the answer in ms when seconds are asked for are the two dominant errors.
3Reading a micrometer for a gamma-absorber thickness (2 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 6 style• micrometer, resolution, absorber, Unit 6
Question
A micrometer screw gauge (thimble graduated in 0.01 mm) with no zero error is used to measure the thickness of a lead absorber sheet. The main scale reads 2.5 mm and the thimble reads 18 divisions. State the thickness to the resolution of the instrument. (2 marks)
Step-by-step solution
1. Step 1
  The thimble adds hundredths of a millimetre: each division is 0.01 mm and it reads 18.
  $\text{thimble} = 18 \times 0.01 = 0.18\ \text{mm}$
2. Step 2
  Add the main-scale and thimble parts. With no zero error, no correction is needed; quote to 2 d.p. (0.01 mm resolution).
  $x = 2.5 + 0.18 = 2.68\ \text{mm}$
Answer
Thickness $= 2.5 + 0.18 = 2.68\ \text{mm}$ (quoted to the 0.01 mm resolution).
Examiner tip
Marks: (1) thimble contribution 0.18 mm added to 2.5 mm; (2) 2.68 mm to 2 d.p. Rounding away the thimble reading (writing 2.7 mm) throws away the resolution mark. Always check for a zero error before recording.
4Background-corrected count rate (3 marks)
Building confidenceMulti-step problem• Edexcel IAL Physics Unit 6 style• count rate, background radiation, Unit 6
Question
With no source present, a GM tube records 90 counts in 60 s (background). With a beta source in place it records 3690 counts in 60 s. Calculate the corrected (source-only) count rate. (3 marks)
Step-by-step solution
1. Step 1
  Find the background count rate first: background counts divided by time.
  $R_{bg} = \frac{90}{60} = 1.5\ \text{s}^{-1}$
2. Step 2
  Find the measured count rate with the source present.
  $R_{meas} = \frac{3690}{60} = 61.5\ \text{s}^{-1}$
3. Step 3
  The corrected count rate is the measured rate minus the background rate.
  $R = R_{meas} - R_{bg} = 61.5 - 1.5 = 60\ \text{s}^{-1}$
Answer
Background $= 1.5\ \text{s}^{-1}$ ; measured $= 61.5\ \text{s}^{-1}$ ; corrected count rate $= 60\ \text{s}^{-1}$ .
Examiner tip
Marks: (1) background rate 1.5 s⁻¹; (2) measured rate 61.5 s⁻¹; (3) subtract to get 60 s⁻¹. The background MUST be subtracted — omitting it leaves a systematic overestimate of the source activity. Subtract rates (not raw counts) unless the counting times are identical.
5Period from timing many oscillations (3 marks)
Building confidenceMulti-step problem• Edexcel IAL Physics Unit 6 style• period, oscillations, random error, Unit 6
Question
To find the period of a mass–spring oscillator, a student uses a digital timer to record the time for 20 complete oscillations as 27.6 s. Calculate (a) the period of one oscillation and (b) the frequency. (3 marks)
Step-by-step solution
1. Step 1
  The period of one oscillation is the total time divided by the number of oscillations timed.
  $T = \frac{t}{n} = \frac{27.6}{20} = 1.38\ \text{s}$
2. Step 2
  Timing many oscillations and dividing spreads the timing (reaction) error over 20 swings, so each period value is far less affected than timing one swing.
  $T = 1.38\ \text{s (per oscillation)}$
3. Step 3
  Frequency is the reciprocal of the period.
  $f = \frac{1}{T} = \frac{1}{1.38} = 0.725\ \text{Hz}$
Answer
(a) $T = 27.6 / 20 = 1.38\ \text{s}$ . (b) $f = 1/1.38 = 0.725\ \text{Hz}$ .
Examiner tip
Marks: (1) divide total time by 20 → 1.38 s; (2) recognise this reduces the timing error per period; (3) f = 1/T = 0.725 Hz. Timing a single oscillation makes the ~0.2 s reaction error a large fraction of the reading; timing 20 and dividing shrinks it. Do NOT forget to divide by the number of oscillations.
6Anomaly versus statistical scatter in counts (3 marks)
Building confidenceMulti-step problem• Edexcel IAL Physics Unit 6 style• anomaly, count rate, statistical scatter, Unit 6
Question
A student records the counts in successive 10 s intervals from a fixed source: 512, 498, 505, 143, 507. (a) Identify the anomalous reading. (b) Explain why the small differences between 498, 505, 507 and 512 are NOT anomalies. (c) Give a sensible mean count for a 10 s interval. (3 marks)
Step-by-step solution
1. Step 1
  An anomaly lies well outside the pattern. Four readings cluster near 500; 143 is far below them, so it is anomalous (e.g. the timer or source was disturbed).
  $\text{anomaly} = 143$
2. Step 2
  Radioactive decay is random, so counts naturally scatter by about $\pm\sqrt{N}\approx\pm 22$ around 500. Differences of a few counts are expected statistical variation, not mistakes.
  $\sqrt{500} \approx 22$
3. Step 3
  Discard the anomaly, then average the four consistent readings.
  $\bar{N} = \frac{512+498+505+507}{4} = \frac{2022}{4} = 505.5 \approx 506$
Answer
(a) 143 is the anomaly. (b) The spread around 500 is statistical scatter of order $\pm\sqrt{N}\approx\pm22$ , so it is expected, not a mistake. (c) Mean of the remaining four $\approx 506$ counts.
Examiner tip
Marks: (1) 143 identified as the anomaly; (2) small spread explained as statistical (±√N) scatter, not an error; (3) mean of the four consistent readings ≈ 506. The examiner discriminator is distinguishing genuine anomalies (discard) from the inherent random scatter of counts (keep).
7Time constant from an oscilloscope discharge trace (3 marks)
StretchGraph or diagram• Edexcel IAL Physics Unit 6 style• oscilloscope, time constant, capacitor discharge, Unit 6
Question
A capacitor discharges through a resistor. On an oscilloscope the p.d. starts at 8.0 V (Y-gain 2.0 V/div) and the trace falls to 0.37 of that value after 2.5 horizontal divisions. The time-base is 10 ms per division. Determine the time constant of the circuit. (3 marks)
Step-by-step solution
1. Step 1
  The time constant $\tau$ is the time for the p.d. to fall to $\tfrac{1}{e} \approx 0.37$ of its initial value. Here the p.d. falls to $0.37\times 8.0 = 3.0$ V (≈ $V_0/e$ ), so the time to reach that level is one time constant.
  $V = \tfrac{V_0}{e} \approx 0.37 V_0 \Rightarrow t = \tau$
2. Step 2
  Read the horizontal interval: 2.5 divisions at 10 ms per division.
  $\tau = (\text{divisions}) \times (\text{time/div}) = 2.5 \times 10\ \text{ms}$
3. Step 3
  Evaluate and convert to seconds.
  $\tau = 25\ \text{ms} = 25 \times 10^{-3} = 0.025\ \text{s}$
Answer
The trace reaches $0.37 V_0$ after 2.5 divisions, so $\tau = 2.5 \times 10 = 25\ \text{ms} = 0.025\ \text{s}$ .
Examiner tip
Marks: (1) recognise the 0.37V₀ (V₀/e) level marks one time constant; (2) read time = divisions × time/div; (3) 25 ms = 0.025 s. The Y-gain (2.0 V/div) is a distractor for the time reading — only the time-base matters for τ. Forgetting the time/div or misreading the 0.37 level are the usual losses.
8Micrometer with a zero error on a wire diameter (3 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 6 style• micrometer, zero error, systematic error, Unit 6
Question
Before measuring the diameter of a spring wire, a micrometer is closed fully and reads −0.02 mm. It then reads 1.24 mm on the wire. State the type of error, explain whether the zero error is added or subtracted, and give the corrected diameter. (3 marks)
Step-by-step solution
1. Step 1
  The −0.02 mm reading with the jaws closed is a zero error, which is a systematic error — it shifts every reading by the same fixed amount.
  $\text{zero error} = -0.02\ \text{mm}$
2. Step 2
  Correct with corrected = reading − zero error. Subtracting a negative value adds 0.02 mm, because the instrument reads 0.02 mm too low.
  $\text{corrected} = 1.24 - (-0.02) = 1.24 + 0.02$
3. Step 3
  Evaluate, keeping the 0.01 mm resolution.
  $d = 1.26\ \text{mm}$
Answer
Zero error $= -0.02\ \text{mm}$ (systematic); corrected $= 1.24 - (-0.02) = 1.26\ \text{mm}$ .
Examiner tip
Marks: (1) zero error = systematic error; (2) a negative zero error is added back (corrected = reading − zero error); (3) 1.26 mm. A zero error is not reduced by repeating readings — it is removed only by re-zeroing or subtracting it. Getting the sign wrong is the classic lost mark.
9Suggesting specific IA2 apparatus improvements (3 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 6 style• improvement, resolution, systematic error, Unit 6
Question
In a radioactivity experiment a student (i) does not subtract the background count, (ii) times the swing of a pendulum by eye for a single oscillation with a stopwatch, and (iii) measures a small absorber thickness with a millimetre ruler. For EACH, name a specific improvement and state precisely why it improves the result. (3 marks)
Step-by-step solution
1. Step 1
  Background not subtracted — remove a systematic error. Record the count with no source present and subtract the background count rate from every measured rate, so the corrected rate reflects the source alone.
2. Step 2
  Single-swing timing — reduce a random (reaction-time) error. Time many oscillations (e.g. 20) with a digital timer / light gates and divide by the number, so the ~0.2 s reaction error is spread over many periods (or removed entirely by light gates).
3. Step 3
  Absorber thickness with a ruler — higher resolution. Use a micrometer screw gauge (0.01 mm resolution) instead of a ruler (1 mm) — 100× finer resolution, greatly reducing the percentage uncertainty in the thickness.
Answer
Radioactivity: measure and subtract the background count rate (removes a systematic overestimate). Timing: time 20 oscillations with a digital timer/light gates and divide (reduces/removes reaction-time error). Thickness: use a micrometer (0.01 mm) not a ruler (1 mm) — far smaller percentage uncertainty.
Examiner tip
Marks: (1) subtract background named + systematic-error benefit; (2) many oscillations / light gates + reaction-time benefit; (3) micrometer + resolution benefit. Each improvement must NAME the technique/instrument AND the specific error it addresses — 'be more careful' is never credited.

Model Answers — Practical Skills II - Implementation and Measurement

High-scoring sample answers for practical skills ii - implementation and measurement on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the count rate of a radioactive source, and calculate it for a scaler that records 6000 counts in 300 s.
Model answer
The count rate is the number of counts registered by the detector per unit time (e.g. counts per second).

$\text{count rate} = \frac{\text{counts}}{\text{time}} = \frac{6000}{300} = 20\ \text{s}^{-1}$

The count rate is 20 s⁻¹ (20 counts per second).
Why this scores
Why this scores 2/2. (1) count rate = counts per unit time; (2) 6000 ÷ 300 = 20 s⁻¹ with the unit. Quoting a bare count (6000) as the 'count rate', or omitting the unit, loses the second mark.
Question 2
2 marks
Q (2 marks). On an oscilloscope, a discharging capacitor's p.d. falls to a chosen level over 4.0 horizontal divisions. The time-base is set to 5.0 ms per division. State the time interval in seconds.
Model answer
The time interval is the number of horizontal divisions times the time-base setting:

$t = 4.0 \times 5.0\ \text{ms} = 20\ \text{ms}$

Converting to seconds ( $1\ \text{ms} = 10^{-3}\ \text{s}$ ):

$t = 20 \times 10^{-3} = 0.020\ \text{s}$
Why this scores
Why this scores 2/2. (1) t = divisions × time/div = 20 ms; (2) converted to 0.020 s. Forgetting the time/div (quoting 4.0 s) or leaving the answer in ms are the usual errors.
Question 3
2 marks
Q (2 marks). A micrometer with no zero error shows a main-scale reading of 3.0 mm and a thimble reading of 27 (thimble graduated in 0.01 mm). State the measured thickness to the resolution of the instrument.
Model answer
The thimble adds $27 \times 0.01 = 0.27\ \text{mm}$ to the main-scale reading:

$\text{reading} = 3.0 + 0.27 = 3.27\ \text{mm}$

The value is 3.27 mm, quoted to 2 decimal places to match the 0.01 mm resolution.
Why this scores
Why this scores 2/2. (1) thimble contribution 0.27 mm added to 3.0 mm; (2) 3.27 mm to the 0.01 mm resolution. Rounding to 3.3 mm throws away the resolution mark.
Question 4
3 marks
Q (3 marks). A GM tube records 240 counts in 120 s with no source present, and 6240 counts in 120 s with a gamma source present. (a) State why the background must be measured. (b) Calculate the corrected count rate of the source.
Model answer
(a) Background radiation (from cosmic rays, rocks, building materials, etc.) is always present and is also detected by the tube. If it is not subtracted, the recorded count rate overestimates the source's activity — a systematic error.

(b) Work in rates:

$R_{bg} = \frac{240}{120} = 2.0\ \text{s}^{-1}, \qquad R_{meas} = \frac{6240}{120} = 52\ \text{s}^{-1}$

$R = R_{meas} - R_{bg} = 52 - 2.0 = 50\ \text{s}^{-1}$

The corrected count rate is 50 s⁻¹.
Why this scores
Why this scores 3/3. (a) background is always present and would inflate the rate (systematic error); (b) background rate 2.0 s⁻¹ and measured 52 s⁻¹, subtracted → 50 s⁻¹. Since the counting times are equal you could subtract counts first, but subtracting rates is always safe.
Question 5
4 marks
Q (4 marks). A capacitor discharges through a resistor and the p.d. is displayed on an oscilloscope. The trace starts at 6.0 V and falls to 2.2 V after 3.0 horizontal divisions. The time-base is 8.0 ms per division. (a) Explain how you can identify one time constant on the trace. (b) Determine the time constant.
Model answer
(a) After one time constant $\tau$ , the p.d. falls to $\tfrac{1}{e} \approx 0.37$ of its initial value. Here $0.37 \times 6.0 = 2.2\ \text{V}$ , so the point where the trace reaches 2.2 V marks one time constant, and the time to reach it (read from the horizontal scale) is $\tau$ .

(b) The time to reach 2.2 V is 3.0 divisions at 8.0 ms/div:

$\tau = 3.0 \times 8.0\ \text{ms} = 24\ \text{ms} = 24 \times 10^{-3} = 0.024\ \text{s}$

The time constant is 0.024 s (24 ms).
Why this scores
Why this scores 4/4. (a) τ is the time to fall to 1/e ≈ 0.37 of V₀ (here 2.2 V); (b) time = divisions × time/div = 24 ms = 0.024 s. Reading the wrong p.d. level (e.g. half of V₀) or forgetting the time/div are the traps.
Question 6
4 marks
Q (4 marks). A student uses light gates and a data-logger to time 25 oscillations of a mass on a spring as 35.0 s. (a) Calculate the period and the frequency. (b) State one advantage of timing 25 oscillations rather than one, and one advantage of light gates over a hand-operated stopwatch.
Model answer
(a) Period of one oscillation:

$T = \frac{35.0}{25} = 1.40\ \text{s}, \qquad f = \frac{1}{T} = \frac{1}{1.40} = 0.714\ \text{Hz}$

(b) Timing 25 oscillations and dividing spreads the timing (reaction) error over 25 periods, so the uncertainty in a single period is much smaller. Light gates with a data-logger remove human reaction time altogether, because the timing is triggered electronically.
Why this scores
Why this scores 4/4. (a) T = 35.0/25 = 1.40 s and f = 1/T = 0.714 Hz; (b) timing many oscillations reduces the reaction error per period, and light gates remove reaction time entirely. Forgetting to divide by 25, or leaving T in the wrong form, are the common slips.
Question 7
4 marks
Q (4 marks). Counts are recorded in five successive 30 s intervals from a fixed source: 1204, 1189, 1211, 655, 1198. (a) Identify the anomaly and give a likely cause. (b) Explain why the differences among 1189, 1198, 1204 and 1211 should NOT be treated as anomalies. (c) Calculate a mean count for a 30 s interval.
Model answer
(a) Four readings cluster near 1200, but 655 lies far below them, so it is the anomaly. A likely cause is a timing/counting fault — e.g. the count was stopped early or the source was momentarily shielded.

(b) Radioactive decay is a random process, so successive counts naturally scatter by about $\pm\sqrt{N} \approx \pm\sqrt{1200} \approx \pm 35$ about the mean. Differences of a few tens of counts are expected statistical variation, not mistakes.

(c) Discarding the anomaly and averaging the four consistent readings:

$\bar{N} = \frac{1204 + 1189 + 1211 + 1198}{4} = \frac{4802}{4} = 1200.5 \approx 1201$
Why this scores
Why this scores 4/4. (a) 655 identified with a sensible cause; (b) small spread explained as statistical (±√N ≈ ±35) scatter of a random process; (c) anomaly excluded, mean ≈ 1201. The discriminator is separating a genuine anomaly (discard) from the inherent random scatter of counts (keep).
Question 8
6 marks
Q (6 marks). In a capacitor-discharge experiment a student times the p.d. falling to half its value with a stopwatch, repeating 10 times and getting values scattered between 4.1 s and 5.3 s. The voltmeter used draws a small current from the circuit. (a) State the type of error the 4.1–5.3 s scatter represents and how to reduce it. (b) Explain how the voltmeter drawing current affects the discharge and what type of error it introduces. (c) Explain why simply repeating the timing more times cannot remove the voltmeter effect.
Model answer
(a) The scatter is a random error. The times vary unpredictably above and below a central value (4.1–5.3 s), which is the signature of random error — largely human reaction time in starting and stopping the stopwatch. It is reduced by repeating and taking a mean (and excluding anomalies), or better by using a data-logger / oscilloscope to time the discharge electronically.

(b) The voltmeter provides an extra discharge path — a systematic error. A voltmeter that draws current lets charge leak away through the meter as well as through the intended resistor, so the capacitor discharges faster than it should. This shifts every timing the same way (times too short), which is a systematic error. It is reduced by using a high-resistance (or digital) voltmeter, or a data-logger with very high input impedance.

(c) Repeating cannot fix a systematic error. Taking more readings only averages out the random scatter and improves the precision of the mean; it does not remove the constant bias caused by the voltmeter's extra current, because every reading is shifted the same way. Only changing to a high-impedance meter (or correcting for it) restores accuracy.
Why this scores
Why this scores 6/6. (a) random error from reaction time + reduced by repeats/mean or electronic timing (2); (b) voltmeter loading gives an extra discharge path → discharges too fast → systematic error, fixed by a high-resistance/digital meter (2); (c) repeating averages out random error only, so the systematic bias survives (2). The key discriminator is stating that a systematic error is unaffected by repetition.
Question 9
6 marks
Q (6 marks). A student investigating the absorption of gamma rays by lead (i) records the count rate for each lead thickness but never measures the background, (ii) measures each lead sheet's thickness with a millimetre ruler, and (iii) counts for only 10 s at each thickness. Identify three weaknesses and give a specific improvement for each.
Model answer
Weakness 1 — background not measured. The recorded count rate includes a background contribution, so the plotted rates are systematically too high and the absorption graph is distorted. Improvement: measure the count rate with no source present and subtract this background count rate from every reading, giving the true source-only rate.

Weakness 2 — thickness measured with a ruler. Thin lead sheets are a fraction of a millimetre thick, far below the ruler's 1 mm resolution, giving a very large percentage uncertainty. Improvement: use a micrometer screw gauge (0.01 mm resolution), measuring at several points and averaging, so the percentage uncertainty in the thickness is much smaller.

Weakness 3 — counting for only 10 s. Short counting times give few counts, so the statistical scatter ( $\pm\sqrt{N}$ ) is a large fraction of each reading. Improvement: count for a much longer time (e.g. several minutes) at each thickness, so $N$ is large and the relative scatter $\sqrt{N}/N = 1/\sqrt{N}$ is small, improving reliability.
Why this scores
Why this scores 6/6. Three named weaknesses each paired with a SPECIFIC improvement and its benefit: subtract background (removes systematic overestimate), micrometer (resolution → smaller % uncertainty), longer counting time (larger N → smaller relative statistical scatter). Each mark pair needs both the technique AND the error it addresses.
Question 10
6 marks
Q (6 marks). A capacitor discharges through a resistor and the p.d. is displayed on an oscilloscope with the time-base at 5.0 ms per division. The trace starts at 8.0 V and the p.d. falls to 3.0 V after 2.6 horizontal divisions. Each horizontal reading can be judged to ±0.2 of a division. (a) Explain how this identifies one time constant, and determine the time constant. (b) Estimate the percentage uncertainty in the time constant from the reading uncertainty. (c) State, with reasons, one change to the apparatus or method that would reduce this percentage uncertainty.
Model answer
(a) After one time constant $\tau$ the p.d. falls to $\tfrac{1}{e} \approx 0.37$ of its initial value: $0.37 \times 8.0 = 3.0\ \text{V}$ . So the point where the trace reaches 3.0 V marks one time constant, and the horizontal time to it is $\tau$ :

$\tau = 2.6 \times 5.0\ \text{ms} = 13\ \text{ms} = 0.013\ \text{s}$

(b) The reading uncertainty is $\pm 0.2$ division on a $2.6$ -division reading, so:

$\%\ \text{unc.} = \frac{0.2}{2.6} \times 100 = 7.7\% \approx 8\%$

giving $\tau = 13 \pm 1\ \text{ms}$ .

(c) Use a faster time-base (fewer ms per division) so the discharge spans more horizontal divisions; the same $\pm 0.2$ -division reading error is then a smaller fraction of a larger division count, cutting the percentage uncertainty. (Alternatively, log the p.d. with a data-logger and plot $\ln V$ against $t$ , using the gradient $-1/(RC)$ over many points to average out individual reading errors.)
Why this scores
Why this scores 6/6. (a) 3.0 V = 0.37V₀ = V₀/e marks one τ, and τ = 2.6 × 5.0 = 13 ms (2); (b) % uncertainty = (0.2/2.6)×100 ≈ 8%, so τ = 13 ± 1 ms (2); (c) a faster time-base spreads the discharge over more divisions so the same ±0.2-division error is a smaller fraction, or a data-logger + ln V–t graph averages many points (2). The discriminator is linking the reading uncertainty to the number of divisions and proposing a change that increases that number.

Key Formulae — Practical Skills II - Implementation and Measurement

The formulae you need to memorise for practical skills ii - implementation and measurement on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Count rate
$\text{count rate} = \frac{\text{counts}}{\text{time}}$
$counts$
number of counts recorded by the scaler
$time$
counting time / s
When to use
Converting a raw count from a GM tube + scaler into a rate. Always state the unit (s⁻¹ or counts per minute). Use the same time base when comparing rates.
Example
4800 counts in 120 s: $\text{count rate} = 4800/120 = 40\ \text{s}^{-1}$ .
Background-corrected count rate
$R = R_{meas} - R_{bg}$
$R$
corrected (source-only) count rate / s⁻¹
$R_{meas}$
measured count rate with the source present
$R_{bg}$
background count rate (no source)
When to use
Whenever a radioactive source's count rate is measured. Subtract the background rate so the result reflects the source alone. Convert both to rates before subtracting if the counting times differ.
Example
Measured 61.5 s⁻¹, background 1.5 s⁻¹: $R = 61.5 - 1.5 = 60\ \text{s}^{-1}$ .
Time interval from an oscilloscope
$t = (\text{no. of divisions}) \times (\text{time/div})$
$t$
time interval read from the screen / s
$no. of divisions$
horizontal divisions spanned
$time/div$
time-base setting (e.g. 10 ms per division)
When to use
Reading any time interval (a period, a discharge time, a time constant) off a CRO or data-logger trace. Convert the time/div to seconds before combining with other formulae.
Example
3.5 divisions at 20 ms/div: $t = 3.5 \times 20 = 70\ \text{ms} = 0.070\ \text{s}$ .
Period from many oscillations
$T = \frac{t}{n}$
$T$
period of one oscillation / s
$t$
total time for n oscillations / s
$n$
number of complete oscillations timed
When to use
Finding a period from a digital timer or light gates. Timing many oscillations and dividing reduces the timing (reaction) error per period.
Example
20 oscillations in 27.6 s: $T = 27.6/20 = 1.38\ \text{s}$ .
Frequency from period
$f = \frac{1}{T}$
$f$
frequency / Hz
$T$
period / s (must be in seconds)
When to use
After finding a period from a timer or oscilloscope. T MUST be in seconds, or the frequency comes out in the wrong units.
Example
$T = 1.38\ \text{s} \Rightarrow f = 1/1.38 = 0.725\ \text{Hz}$ .
Uncertainty of a single analogue reading
$\Delta x = \tfrac{1}{2}\times(\text{smallest division})$
$\Delta x$
absolute uncertainty of one reading
$smallest division$
the resolution of the instrument scale
When to use
Quoting the uncertainty of a single reading on an analogue scale (ruler, thermometer, dial). For a digital instrument the uncertainty is ± the last displayed digit.
Example
Micrometer with 0.01 mm divisions: $\Delta x = \tfrac{1}{2}\times 0.01 = \pm 0.005\ \text{mm}$ .

Key Definitions and Keywords — Practical Skills II - Implementation and Measurement

Definitions to memorise and the exact keywords mark schemes credit for practical skills ii - implementation and measurement answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Resolution
Examiner keyword
The smallest change in a measured quantity that an instrument can detect and display — usually equal to its smallest scale division (e.g. 0.01 mm for a micrometer, or one time/div step on an oscilloscope).
Zero error
Examiner keyword
A non-zero reading given by an instrument when the true value should be zero (e.g. a micrometer reading −0.02 mm with the jaws closed). It is a systematic error and must be subtracted from every reading (corrected = reading − zero error).
Parallax error
Examiner keyword
An error caused by reading a scale from the wrong angle, so the apparent position of the pointer/liquid/trace relative to the scale is shifted. Avoided by viewing perpendicular to the scale (eye directly above, or using a mirror/set-square).
Anomaly (anomalous result)
Examiner keyword
A reading that lies well outside the pattern of the other repeat readings, usually due to a mistake. It is identified and excluded before calculating a mean — distinct from the expected statistical scatter of radioactive counts.
Precision
Examiner keyword
How close repeated readings are to one another (the spread/repeatability). Precise readings have a small range. Precision is independent of accuracy.
Accuracy
Examiner keyword
How close a measured value (usually a mean) is to the true value. A result can be precise but not accurate if a systematic error (e.g. an unsubtracted background) shifts every reading.
Systematic error
Examiner keyword
An error that shifts every reading by the same amount or in the same direction (e.g. zero error, an unsubtracted background count, a voltmeter loading a circuit). It is NOT reduced by repeating readings; it is removed by re-zeroing, recalibrating, or correcting for it.
Random error
Examiner keyword
An error that varies unpredictably from reading to reading (e.g. reaction time in timing, the random scatter of radioactive counts). Reduced by repeating readings and taking a mean, or by counting for longer.
Background radiation / background count
Examiner keyword
The ionising radiation always present from natural and artificial sources (cosmic rays, rocks, building materials, etc.). Its count rate is measured with no source present and subtracted from every measured count rate.
Count rate
Examiner keyword
The number of counts registered by a detector per unit time (counts ÷ time), e.g. in s⁻¹ or counts per minute. The corrected count rate is the measured rate minus the background rate.
Time constant (τ)
Examiner keyword
For a capacitor discharging through a resistor, the time for the p.d. (or charge/current) to fall to 1/e ≈ 0.37 of its initial value. On an oscilloscope trace, the time from the start to the 0.37V₀ level, read as divisions × time/div.
Significant figures
Examiner keyword
The digits in a value that carry meaning about its precision. Readings from one instrument should be recorded to a consistent number of significant figures (or decimal places), reflecting the instrument's resolution; ln/log columns are typically quoted to 3 s.f.
Range
Examiner keyword
The difference between the largest and smallest of a set of repeat readings. Half the range is often used as an estimate of the uncertainty in the mean.
Time-base (time/div)
Examiner keyword
The oscilloscope setting that sets how much time each horizontal division represents (e.g. 10 ms per division). A time interval is the number of horizontal divisions × the time/div.
Statistical scatter of counts
Examiner keyword
The natural variation in successive radioactive counts because decay is random. For a count N the standard spread is about ±√N, so small differences between repeat counts are expected, not anomalies.

Common Mistakes and Misconceptions — Practical Skills II - Implementation and Measurement

The traps other students keep falling into on practical skills ii - implementation and measurement questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Not subtracting the background count from a measured count rate
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students record the count with the source present and quote it directly, forgetting that background radiation is always being detected too.
How to avoid it
Always measure the background count rate with no source present and subtract it: corrected rate = measured rate − background rate. Omitting it is a systematic error that inflates the source activity.
✕Forgetting to multiply the number of divisions by the time/div when reading a time off an oscilloscope
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students read the number of divisions and treat it directly as the time, ignoring the time-base setting.
How to avoid it
A time interval is t = (horizontal divisions) × (time/div). Read the time-base setting off the oscilloscope and multiply by it every time, then convert to seconds.
✕Timing a single oscillation instead of many, then dividing
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students start and stop the timer on one swing, so the ~0.2 s reaction error is a large fraction of the period.
How to avoid it
Time many oscillations (e.g. 20) and divide by the number: T = t / n. This spreads the timing error over all the swings (or use light gates to remove reaction time).
✕Recording readings to the wrong (or inconsistent) number of significant figures
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students copy whatever the display shows without matching it to the instrument's resolution, mixing precisions within a table; ln/log columns are rounded inconsistently.
How to avoid it
Record every reading from one instrument to a consistent number of significant figures set by its resolution (e.g. 2 d.p. in mm for a micrometer), and quote ln/log columns to 3 s.f. Do not over-state precision the scale cannot support.
✕Ignoring a zero error, or getting its sign wrong (adding when you should subtract)
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students record the raw reading without checking the closed-jaw value, or forget that a negative zero error is added back.
How to avoid it
Always check the reading with the instrument set to zero first. Then corrected = reading − zero error: a positive zero error is subtracted, a negative one is added back. A zero error is systematic — repeating readings will not remove it.
✕Treating the natural scatter of radioactive counts as a mistake (or as anomalies to be discarded)
Edexcel IAL Physics Unit 6 examiner reports
Why it happens
Students expect repeat counts to be identical and label every small difference as an error, not realising decay is a random process.
How to avoid it
Remember counts scatter by about ±√N because decay is random — small differences are expected statistical variation, not anomalies. Only discard a reading that lies far outside this spread, and count for longer (larger N) to reduce the relative scatter.

Practical Skills II - Implementation and Measurement — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Practical Skills II - Implementation and Measurement

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Count rate from a scaler over a fixed time (2 marks)

2Reading a time interval off an oscilloscope trace (2 marks)

3Reading a micrometer for a gamma-absorber thickness (2 marks)

4Background-corrected count rate (3 marks)

5Period from timing many oscillations (3 marks)

6Anomaly versus statistical scatter in counts (3 marks)

7Time constant from an oscilloscope discharge trace (3 marks)

8Micrometer with a zero error on a wire diameter (3 marks)

9Suggesting specific IA2 apparatus improvements (3 marks)

Count rate

Background-corrected count rate

Time interval from an oscilloscope

Period from many oscillations

Frequency from period

Uncertainty of a single analogue reading

Resolution

Zero error

Parallax error

Anomaly (anomalous result)

Precision

Accuracy

Systematic error

Random error

Background radiation / background count

Count rate

Time constant (τ)

Significant figures

Range

Time-base (time/div)

Statistical scatter of counts

✕Not subtracting the background count from a measured count rate

✕Forgetting to multiply the number of divisions by the time/div when reading a time off an oscilloscope

✕Timing a single oscillation instead of many, then dividing

✕Recording readings to the wrong (or inconsistent) number of significant figures

✕Ignoring a zero error, or getting its sign wrong (adding when you should subtract)

✕Treating the natural scatter of radioactive counts as a mistake (or as anomalies to be discarded)

Practical Skills II - Implementation and Measurement — frequently asked questions