What are the key steps involved in processing results in Practical Skills II for Edexcel International A Levels Physics?

In Practical Skills II for Edexcel International A Levels Physics, processing results involves several key steps: collecting data accurately, organizing data into tables or charts, analyzing data to identify patterns or trends, calculating uncertainties, and interpreting the results in the context of the experiment. Each step is crucial for ensuring the reliability and validity of the experimental findings.

How do I calculate uncertainties in my experimental results for Edexcel International A Levels Physics?

Calculating uncertainties in experimental results involves determining the precision of your measurements and considering any potential sources of error. For Edexcel International A Levels Physics, this typically includes using the smallest division of your measuring instrument as a basic uncertainty, applying percentage uncertainty for calculated values, and combining uncertainties using appropriate mathematical methods. Understanding and calculating uncertainties is essential for accurately interpreting your experimental results.

Why is data analysis important in Practical Skills II - Processing Results?

Data analysis is a critical component of Practical Skills II - Processing Results because it allows students to make sense of the raw data collected during experiments. Through analysis, students can identify trends, test hypotheses, and draw meaningful conclusions. This process helps in validating the experimental procedure and ensuring that the results are reliable and applicable to real-world scenarios, which is a key learning outcome for Edexcel International A Levels Physics.

What are some common methods for presenting data in Practical Skills II for Edexcel International A Levels Physics?

Common methods for presenting data in Practical Skills II include using tables, graphs, and charts. Tables are useful for organizing numerical data clearly, while graphs such as line graphs, bar charts, and scatter plots help visualize relationships and trends. Choosing the appropriate method depends on the type of data and the information you wish to convey. Effective data presentation is crucial for communicating your findings clearly and accurately.

How can I ensure the accuracy and reliability of my results in Practical Skills II - Processing Results?

To ensure accuracy and reliability in Practical Skills II - Processing Results, it is important to follow a systematic approach: use calibrated instruments, repeat measurements to identify anomalies, apply consistent methods, and document all procedures meticulously. Additionally, understanding and accounting for uncertainties, as well as critically evaluating the experimental design, can help improve the reliability of your results. These practices are essential for achieving high standards in Edexcel International A Levels Physics.

Why is "Confusing ln (base e) with log (base 10) when linearising — e.g. plotting lg for an exponential" a common mistake in Practical Skills II - Processing Results?

Why it happens: Both ‘log’ buttons exist on the calculator and the difference in method is not memorised. How to avoid it: Use **ln (base e) for exponentials** y = y_0e^-kx (plot y vs x, gradient = -k), and **lg (base 10) for power laws** y = Ax^n (plot y vs x, gradient = n). Mixing them scales the gradient by 10 2.303. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Why is "Forgetting the minus sign when reading $\lambda$ or $1/RC$ from the gradient of a $\ln$ plot" a common mistake in Practical Skills II - Processing Results?

Why it happens: Students read the gradient's magnitude but overlook that the decay gradient is negative, so = -gradient. How to avoid it: Write the linearised form first: A = A_0 - t, so the **gradient is -** (negative). Therefore = -gradient, a **positive** value. The same applies to -1/RC and -. A negative or RC is a sure sign of this error. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Why is "Not multiplying the percentage uncertainty by the power for a squared (or cubed) term" a common mistake in Practical Skills II - Processing Results?

Why it happens: Students add each measurement's percentage uncertainty once, ignoring that some are raised to a power. How to avoid it: For a term x^n the percentage uncertainty is |n|×\%x. A **squared** quantity (e.g. T^2 in k = 4^2m/T^2, or A d^2) contributes **twice** its percentage uncertainty; a cube contributes three times. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Why is "Plotting the raw data (e.g. $A$ vs $t$, a curve) instead of the linearised quantities" a common mistake in Practical Skills II - Processing Results?

Why it happens: Students plot what was measured directly rather than rearranging to the straight-line form first. How to avoid it: Decide the linearisation **before** plotting: for an exponential plot ** y vs x**; for a power law plot ** y vs x**. A curved raw graph cannot give a gradient equal to -, -1/RC or n — only the linearised straight line can. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Why is "Using a small gradient triangle (a fraction of the line) to read the gradient" a common mistake in Practical Skills II - Processing Results?

Why it happens: Students pick two convenient nearby grid points instead of two well-separated points on the line. How to avoid it: Draw a **large triangle spanning at least half the plotted line** and read the coordinates to the scale's precision. A big triangle reduces the fractional error in x and y and protects the accuracy mark. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Why is "Mixing up validity and reliability — e.g. saying ‘taking more counts makes the result more valid’" a common mistake in Practical Skills II - Processing Results?

Why it happens: The two evaluation terms sound similar and are used loosely in everyday speech. How to avoid it: **Reliability** = consistent, repeatable results (improved by repeating/averaging or more counts, which cut random/counting error). **Validity** = measuring the right thing in a fair test (improved by removing systematic error, e.g. subtracting background). More counts improves reliability, NOT validity. Source: Edexcel IAL Physics Unit 6 (WPH16) examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Practical Skills II - Processing Results

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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 — Processing Results — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 6, WPH16)

The IA2 processing-results toolkit for the Unit 6 written practical paper (WPH16), drawing on Units 4 and 5: plotting and reading graphs (sensible scales, single line of best fit, large gradient triangle with units); the core Unit-6 skill of LINEARISING exponential, logarithmic and power-law relationships (plot ln y vs x for capacitor discharge, radioactive decay and gamma absorption; plot log y vs log x to confirm a power or inverse-square law); combining percentage uncertainties (doubling for squared terms) and counting-statistics uncertainty (√N); estimating gradient uncertainty by the max/min-line method; and evaluating experiments (systematic vs random error, validity vs reliability, targeted improvements).

At a glance

Independent variable on the x-axis, dependent on the y-axis. Choose scales that fill at least half the grid each way in easy steps (1/2/5 per square). Draw a single line of best fit, never dot-to-dot.
Gradient from a LARGE triangle (≥ half the line): $m = \dfrac{\Delta y}{\Delta x}$ . Always state the units (y-units ÷ x-units) — the gradient usually equals a physical quantity.
Exponential → ln plot. $y = y_{0}e^{-kx}$ becomes $\ln y = \ln y_{0} - kx$ : plot $\ln y$ vs $x$ , gradient $= -k$ , intercept $= \ln y_{0}$ .
Three key exponentials: capacitor $Q = Q_{0}e^{-t/RC}$ (gradient $-1/RC$ ); decay $A = A_{0}e^{-\lambda t}$ (gradient $-\lambda$ ); absorption $I = I_{0}e^{-\mu x}$ (gradient $-\mu$ ). Use ln, not lg.
Power law → log–log plot. $y = Ax^{n}$ becomes $\lg y = n\lg x + \lg A$ : plot $\lg y$ vs $\lg x$ , gradient $= n$ . A gradient of −2 confirms an inverse-square law; alternatively plot $I$ vs $1/r^{2}$ (straight line through origin).
Percentage uncertainties: ADD for products/quotients, MULTIPLY by the power for a raised term — a squared term ( $T^{2}$ , $d^{2}$ ) contributes twice its %uncertainty.
Counting uncertainty: for a count $N$ , $\Delta N = \sqrt{N}$ , so $\%N = 100/\sqrt{N}$ — record more counts to reduce it. Subtract background first.
Gradient uncertainty (max/min lines): $\Delta m = \dfrac{m_{\text{max}} - m_{\text{min}}}{2}$ . Systematic errors shift accuracy (not cured by repeats); random errors scatter and cut reliability (cured by repeating/averaging).

What you’ll learn

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

Plot experimental data with sensible axes and scales, drawing a single line (or curve) of best fit and identifying anomalies.
Determine a gradient and intercept from a linear graph using a large triangle, stating the units and relating them to a physical quantity.
Linearise an exponential relationship by plotting $\ln y$ against $x$ (capacitor discharge, radioactive decay, gamma absorption) and obtain $-1/RC$ , $-\lambda$ or $-\mu$ from the gradient.
Linearise a power law by plotting $\lg y$ against $\lg x$ , obtain the power $n$ from the gradient, and confirm an inverse-square law (gradient $= -2$ , or $I$ vs $1/r^{2}$ linear through the origin).
Calculate and combine absolute, percentage and counting-statistics ( $\sqrt{N}$ ) uncertainties, doubling the percentage for a squared term, and estimate a gradient's uncertainty by the max/min-line method.
Evaluate an experiment: distinguish systematic from random error and validity from reliability, and suggest specific, targeted improvements.

Graph discipline — scales, best-fit lines and gradients

Fill the grid, draw one best-fit line, and take the gradient from a large triangle with units.

Unit 6 (WPH16) is a written practical paper: you are given data from experiments drawn on Units 4 and 5 and must process them. The marks for good graph technique are the same at IA2 as at IA1, so bank them.

Choosing the axes and scales

Put the independent variable (what the experimenter set — often $t$ or $r$ ) on the x-axis, and the dependent or calculated quantity (such as $\ln A$ ) on the y-axis.
Choose scales so the plotted points fill at least half the grid in each direction, using easy values per square (multiples of 1, 2 or 5 — or 0.25 on a log axis). Awkward scales (3 or 7 per square) cause plotting errors.
For a calculated log quantity ( $\ln V$ , $\lg I$ ) the axis need not start at zero — start just below the smallest value so the points spread out.

Drawing the line

Draw a single, thin line of best fit with roughly as many points above as below. Never join the points dot-to-dot.
On a linearised (ln or log) plot the trend should be a genuine straight line; visible curvature means either the assumed relationship is wrong or a value was mis-calculated.
An anomaly (a point clearly off the trend) is ignored when drawing the best-fit line and checked by repeating — it must not pull the line towards it.

Reading the gradient

Use a large triangle spanning at least half the line, read the two coordinates carefully, and compute $m = \dfrac{\Delta y}{\Delta x}$ .
Always state the units: (y-axis units) ÷ (x-axis units). The gradient usually equals a physical quantity — capacitance on a $Q$ – $V$ plot, $-\lambda$ on a $\ln A$ – $t$ plot, the power $n$ on a log–log plot.

A large gradient triangle spanning most of the line gives a precise gradient. For a Q–V graph the gradient equals the capacitance C. Always quote the gradient's units (here µC V⁻¹ = µF).

IV on x-axis, dependent/calculated quantity on y-axis; fill ≥ half the grid with easy scales.
Draw one thin line of best fit — never dot-to-dot; ignore anomalies when drawing it.
Gradient from a large triangle (≥ half the line); always state units.
The gradient usually equals a physical quantity (C, $-\lambda$ , $n$ , …).

See the full worked example for practical skills ii - processing results →

Linearising exponentials — the ln plot

Take ln of both sides: plot ln y vs x, gradient = −k (so −1/RC, −λ or −μ).

The distinctive IA2 skill is turning a curved exponential decay into a straight line so you can read a constant from the gradient. Take natural logs of both sides — because the base is $e$ , you must use ln, not lg.

For any exponential of the form $y = y_{0}e^{-kx}$ : $\ln y = \ln y_{0} - kx$

Compare with $y = mx + c$ : plot $\ln y$ (y-axis) against $x$ (x-axis). The gradient is $-k$ (negative for decay) and the intercept is $\ln y_{0}$ . Three exam favourites use exactly this:

Relationship	Plot	Gradient	From Unit
Capacitor discharge $Q = Q_{0}e^{-t/RC}$	$\ln Q$ (or $\ln V$ , $\ln I$ ) vs $t$	$-\dfrac{1}{RC}$	Unit 4
Radioactive decay $A = A_{0}e^{-\lambda t}$	$\ln A$ vs $t$	$-\lambda$	Unit 5
Gamma absorption $I = I_{0}e^{-\mu x}$	$\ln I$ vs $x$	$-\mu$	Unit 5

The method

Write the exponential and take ln of both sides to get the $\ln y = \ln y_{0} - kx$ form.
Build a table of $\ln y$ values (calculate to 3 s.f.; be careful to take ln, not lg).
Plot $\ln y$ against $x$ , draw the best-fit straight line, and read the gradient from a large triangle.
The gradient is negative; the constant you want is minus the gradient ( $k = -\text{gradient}$ ), a positive value. For $RC$ use $RC = -\dfrac{1}{\text{gradient}}$ .

Plotting ln A against t linearises the exponential decay A = A₀e^(−λt). The line slopes downward, so its gradient is negative and equals −λ; the decay constant is λ = −gradient, a positive value. The same shape gives −1/RC for a capacitor and −µ for gamma absorption.

Watch the sign. The gradient of a decay ln plot is negative. The physical constant ( $\lambda$ , $1/RC$ , $\mu$ ) is positive, equal to minus the gradient. A negative $\lambda$ or $RC$ is a giveaway that the minus sign was dropped.

Exponential $y = y_{0}e^{-kx}$ → $\ln y = \ln y_{0} - kx$ ; plot $\ln y$ vs $x$ .
Gradient $= -k$ (negative for decay); intercept $= \ln y_{0}$ .
Capacitor: gradient $-1/RC$ ; decay: $-\lambda$ ; absorption: $-\mu$ .
Use ln (base $e$ ), never lg; the constant is $-\text{gradient}$ (positive).

See the full worked example for practical skills ii - processing results →

Linearising power laws — the log–log plot and inverse-square

Take lg of both sides: plot lg y vs lg x, gradient = n; a gradient of −2 confirms inverse square.

When the relationship is a power law $y = Ax^{n}$ (a variable raised to an unknown power, not an exponential), use a log–log plot. Take logs (base 10) of both sides: $\lg y = \lg(Ax^{n}) = \lg A + n\lg x$

Compare with $y = mx + c$ : plot $\lg y$ (y-axis) against $\lg x$ (x-axis). The gradient is the power $n$ and the intercept is $\lg A$ (so $A = 10^{\text{intercept}}$ ).

Confirming an inverse-square law. Radiation from a point source obeys $I = k\,r^{n}$ ; the inverse-square law is the special case $n = -2$ . A log–log plot of $\lg I$ against $\lg r$ should be a straight line of gradient $-2$ . Reading the gradient directly tests the power — you do not have to assume it in advance.

A direct alternative avoids logs entirely: plot $I$ against $1/r^{2}$ . If the inverse-square law holds, this is a straight line through the origin whose gradient is the constant $k$ .

On a log–log plot of lg I against lg r, a power law gives a straight line whose gradient equals the power n. A gradient of −2 (a run of 1 giving a fall of 2 in log units) confirms the inverse-square law I ∝ 1/r². Alternatively, I plotted against 1/r² is a straight line through the origin.

Choosing ln vs lg. The single most important decision:

Exponential ( $y = y_{0}e^{-kx}$ , the variable in the exponent) → ln, plot $\ln y$ vs $x$ , gradient $= -k$ .
Power law ( $y = Ax^{n}$ , the variable raised to a power) → lg (log–log), plot $\lg y$ vs $\lg x$ , gradient $= n$ .

Power law $y = Ax^{n}$ → $\lg y = n\lg x + \lg A$ ; plot $\lg y$ vs $\lg x$ .
Gradient $= n$ (the power); intercept $= \lg A$ , so $A = 10^{\text{intercept}}$ .
Inverse-square: gradient $= -2$ confirms $I \propto 1/r^{2}$ .
Direct test: $I$ vs $1/r^{2}$ is a straight line through the origin.

See the full worked example for practical skills ii - processing results →

Uncertainties — percentages, powers and counting statistics

Add % for products/quotients, multiply by the power for a squared term, and use √N for counts.

Every processed result needs an uncertainty. Work in percentages so uncertainties in different quantities can be combined.

Absolute uncertainty

Single reading: $\Delta x = \tfrac{1}{2}\times$ resolution (a metre rule to 1 mm gives $\pm 0.5\ \text{mm}$ ; a micrometer to 0.01 mm gives $\pm 0.005\ \text{mm}$ ).
Repeated readings: $\Delta x = \dfrac{x_{\text{max}} - x_{\text{min}}}{2}$ (half the spread).

Percentage uncertainty and combining $\%\,x = \frac{\Delta x}{x}\times100\%$

Products and quotients — ADD the percentage uncertainties. For $\rho = RA/L$ , $\%\rho = \%R + \%A + \%L$ .
Powers — MULTIPLY the percentage uncertainty by the power. For $Q = x^{n}$ , $\%Q = |n|\times\%x$ . A squared term (e.g. $T^{2}$ in $k = 4\pi^{2}m/T^{2}$ , or $A \propto d^{2}$ ) contributes twice its percentage uncertainty; a cube contributes three times. Constants ( $4\pi^{2}$ , $\tfrac{4}{3}\pi$ ) have no uncertainty.
A total percentage uncertainty above ~5% suggests significant error worth reducing; identify the dominant term (usually the one with the biggest %, especially after a power).

Counting-statistics uncertainty (radioactivity) A radioactive count $N$ obeys counting statistics: the absolute uncertainty is $\Delta N = \sqrt{N}$ , so: $\%\,N = \frac{\sqrt{N}}{N}\times100 = \frac{100}{\sqrt{N}}$ The percentage uncertainty falls as $1/\sqrt{N}$ , so recording more counts (longer count time) improves precision — quadrupling the count halves the % uncertainty. Always subtract the background count first.

Uncertainty in a gradient — the max/min-line method Draw the steepest and shallowest straight lines that still pass through (most of) the error bars; their gradients bracket the best-fit gradient. Then: $\Delta m = \frac{m_{\text{max}} - m_{\text{min}}}{2}, \qquad \%\,m = \frac{\Delta m}{m_{\text{best}}}\times100$ Use magnitudes when the gradient is negative (as on a decay ln plot).

The steepest (max) and shallowest (min) lines that still pass through the error bars bracket the best-fit line. Half the difference of their gradients is the absolute uncertainty in the gradient, Δm = (m_max − m_min)/2; dividing by the best-fit gradient gives the percentage uncertainty.

Single reading: $\Delta x = \tfrac12\times$ resolution; repeats: $\Delta x = (x_{\max}-x_{\min})/2$ .
Products/quotients: ADD %; powers: MULTIPLY % by the power (square → ×2).
Counts: $\Delta N = \sqrt{N}$ , $\%N = 100/\sqrt{N}$ ; subtract background first.
Gradient uncertainty: $\Delta m = (m_{\max}-m_{\min})/2$ ; divide by $m_{\text{best}}$ for %.

See the full worked example for practical skills ii - processing results →

Evaluating experiments — errors, validity and improvements

Classify the error, decide whether it hurts accuracy or reliability, then propose a targeted fix.

The final marks reward evaluation — judging the experiment and suggesting improvements. Two distinctions are examined again and again.

Systematic vs random error

A systematic error shifts every reading by the same amount in the same direction, so all results are too high or too low. Examples: a zero error, un-subtracted background in a count, voltmeter loading across an $RC$ circuit, or consistent parallax. It reduces accuracy and is not removed by repeating — you must find and correct its cause.
A random error makes readings scatter unpredictably about the true value. Examples: count fluctuations (counting statistics), reaction-time in timing, small draughts. It reduces precision/reliability and is reduced by repeating and averaging (or, for counts, recording more counts).

Validity vs reliability (the classic mix-up)

Validity = whether the experiment measures what it is meant to — a fair test with variables controlled and systematic errors removed (e.g. background subtracted). An invalid experiment gives the wrong answer no matter how often it is repeated.
Reliability = whether the results are consistent and repeatable — small random scatter. Repeating readings (or taking more counts) improves reliability, not validity.

Suggesting improvements An improvement mark needs a specific, feasible change targeting a named error, with the mechanism stated:

Timing dominated by reaction time → time more oscillations (spreads the fixed error, lowering $\%T$ ) or use light gates (removes reaction time).
Large counting uncertainty → count for longer (raises $N$ , so $\%N = 100/\sqrt{N}$ falls), especially at large distances.
Un-subtracted background flattening an inverse-square gradient → measure and subtract background before taking logs.

Vague answers ("be more careful", "use better apparatus") score nothing — the change must be specific and its effect explained.

Systematic: same-direction offset (zero error, background, loading); hurts accuracy; not cured by repeats.
Random: unpredictable scatter (counts, reaction time); hurts reliability; cured by repeating/averaging.
Validity = measures the right thing (remove systematic error); reliability = repeatable (reduce random error).
Improvements must be specific, feasible, and target a named error with its mechanism.

See the full worked example for practical skills ii - processing results →

Quick recap

Put the independent variable on the x-axis, fill ≥ half the grid with easy scales, draw one line of best fit (not dot-to-dot), and take the gradient from a large triangle with units.
Exponential $y = y_{0}e^{-kx}$ : plot $\ln y$ vs $x$ ; gradient $= -k$ (so $-1/RC$ for a capacitor, $-\lambda$ for decay, $-\mu$ for absorption); use ln, not lg.
Power law $y = Ax^{n}$ : plot $\lg y$ vs $\lg x$ ; gradient $= n$ . A gradient of $-2$ confirms an inverse-square law (or plot $I$ vs $1/r^{2}$ , straight line through origin).
Combine uncertainties: ADD % for products/quotients, MULTIPLY % by the power for a raised term (squared → ×2); constants have no uncertainty.
Counting uncertainty: $\Delta N = \sqrt{N}$ , $\%N = 100/\sqrt{N}$ ; subtract background and record more counts to reduce it.
Gradient uncertainty: $\Delta m = (m_{\max}-m_{\min})/2$ from the max/min lines; when a result depends on a single power of the gradient, its % uncertainty equals the gradient's.
Evaluate: systematic error hurts accuracy (not cured by repeats); random error hurts reliability (cured by repeating/more counts); validity ≠ reliability; improvements must be specific and target a named error.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Exponential $y = y_{0}e^{-kx}$ → plot $\ln y$ vs $x$ ; gradient $= -k$ , intercept $= \ln y_{0}$ (use ln, NOT lg).
Capacitor discharge: plot $\ln Q$ (or $\ln V$ ) vs $t$ ; gradient $= -1/RC$ .
Radioactive decay: plot $\ln A$ vs $t$ ; gradient $= -\lambda$ , so $\lambda = -\text{gradient}$ (positive).
Gamma absorption: plot $\ln I$ vs $x$ ; gradient $= -\mu$ .
Power law $y = Ax^{n}$ → plot $\lg y$ vs $\lg x$ ; gradient $= n$ ; gradient $= -2$ confirms $I \propto 1/r^{2}$ .
Percentage uncertainty: $\%x = \dfrac{\Delta x}{x}\times100$ ; ADD for products/quotients, ×power for a raised term (square → ×2).
Counting uncertainty: $\Delta N = \sqrt{N}$ , $\%N = \dfrac{100}{\sqrt{N}}$ ; subtract background first.
Gradient uncertainty: $\Delta m = \dfrac{m_{\max} - m_{\min}}{2}$ from the max/min lines.

How it’s examined

Processing Results is examined on the Unit 6 written practical paper (WPH16, IA2), a structured data-and-methods paper drawing on Units 4 and 5. Typical formats: state what to plot to linearise a given relationship and why (2–3 marks, AO2) — most often an exponential (capacitor discharge, radioactive decay, gamma absorption) needing an $\ln y$ vs $x$ plot, or a power/inverse-square law needing a $\lg y$ vs $\lg x$ plot; determine a gradient from a large triangle with units and relate it to a constant ( $-\lambda$ , $-1/RC$ , $-\mu$ , $n$ ) (3–4 marks); calculate and combine percentage uncertainties, doubling for a squared term, or find a counting uncertainty from $\sqrt{N}$ (3–4 marks); estimate a gradient uncertainty by the max/min-line method (2–3 marks); and an extended evaluation identifying a systematic vs random error and proposing specific, targeted improvements (4–6 marks, AO3). ALWAYS show the linearised form, the substitution with units, and a 3 s.f. answer; keep the minus sign on a decay gradient; and distinguish validity from reliability in evaluations.

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

Take this whole topic with you

Step-by-step worked examples — Practical Skills II - Processing Results

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

Question type:

1Choosing sensible scales for a ln–linear plot (3 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 6 (WPH16) style• graph plotting, scales, ln plot, Unit 6
Question
A student discharges a capacitor and calculates $\ln V$ for each voltage, obtaining values that run from $\ln V = 2.30$ at $t = 0$ down to $\ln V = 0.30$ at $t = 100\ \text{s}$ . The grid provided is 12 large squares wide and 10 large squares tall. State two rules for choosing axes on a results graph, and suggest a suitable scale for the $\ln V$ axis. (3 marks)
Step-by-step solution
1. Step 1
  The independent variable (time $t$ , which the student sets) goes on the x-axis; the calculated dependent quantity ( $\ln V$ ) goes on the y-axis.
2. Step 2
  The plotted points must fill at least half of the grid in each direction, and the scale must be easy to read — use multiples of 1, 2 or 5 per square, never awkward values like 3 or 7.
3. Step 3
  $\ln V$ runs from 0.30 to 2.30, a range of 2.00 over 10 squares. A scale of 0.25 per large square covers 0 to 2.5 and uses about 8 of the 10 squares — more than half, with an easy value per square. (You do NOT have to start the axis at zero for a calculated log quantity.)
  $\ln V\text{ axis: } 0.25\text{ per square} \;\Rightarrow\; 0.0\text{–}2.5\text{ over }10\text{ squares}$
Answer
Independent variable ( $t$ ) on the x-axis, calculated $\ln V$ on the y-axis; points must occupy at least half the grid each way with an easy-to-read scale. A scale of 0.25 per large square is suitable (range 2.00 fills ~8 of the 10 squares).
Examiner tip
Marks: (1) IV ( $t$ ) on x, calculated $\ln V$ on y; (2) scale fills ≥ half the grid AND is a sensible multiple (1/2/5, or 0.25); (3) a workable value such as 0.25 per square. For log/ln axes examiners accept a non-zero start, but the points must still span more than half the grid — a scale that crams them into a corner loses the mark.
2Line of best fit on a linearised plot (2 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 6 (WPH16) style• line of best fit, graph plotting, Unit 6
Question
After linearising an exponential decay, a student plots $\ln A$ against $t$ and the points show a clear straight-line trend with some scatter. Describe how the student should draw the trend line, and state one thing they must not do. (2 marks)
Step-by-step solution
1. Step 1
  Draw a single straight line of best fit — a thin line with roughly as many points above it as below it, balancing the scatter. Once the data are linearised, the trend should genuinely be a straight line.
2. Step 2
  The student must not join the points dot-to-dot (a zig-zag), and must not force the line through the first and last points only. A single straight best-fit line averages out the random scatter and lets a reliable gradient (here $-\lambda$ ) be read.
Answer
Draw one thin, straight best-fit line with a balanced number of points on each side; do NOT connect the points dot-to-dot or force it through the end points only.
Examiner tip
Marks: (1) single straight best-fit line balancing points above/below; (2) explicitly not dot-to-dot. On a linearised (ln or log) plot the whole point is that a straight line should appear — a curved or kinked best-fit line suggests the linearisation is wrong or the data mis-calculated.
3Gradient with a large triangle and its units (3 marks)
Getting startedGraph or diagram• Edexcel IAL Physics Unit 6 (WPH16) style• gradient, units, Unit 6
Question
On a graph of charge $Q$ (in $\mu\text{C}$ ) against potential difference $V$ (in V) for a capacitor, a straight best-fit line through the origin passes through $(0,0)$ and $(8.0\ \text{V},\ 96\ \mu\text{C})$ , which lies near the far end of the line. Determine the gradient, including its units, and state what the gradient represents. (3 marks)
Step-by-step solution
1. Step 1
  Use a large triangle spanning most of the line — here from the origin to the far point $(8.0\ \text{V},\ 96\ \mu\text{C})$ .
  $\text{gradient} = \frac{\Delta Q}{\Delta V} = \frac{96 - 0}{8.0 - 0}$
2. Step 2
  Evaluate the gradient.
  $\text{gradient} = \frac{96}{8.0} = 12\ \mu\text{C V}^{-1}$
3. Step 3
  The units of the gradient are (y-units) ÷ (x-units) $= \mu\text{C} \div \text{V} = \mu\text{F}$ . For a $Q$ – $V$ graph the gradient is $\dfrac{Q}{V} = C$ , the capacitance.
  $\text{gradient} = 12\ \mu\text{F} = C \;(\text{the capacitance})$
Answer
Gradient $= 12\ \mu\text{C V}^{-1} = 12\ \mu\text{F}$ ; it represents the capacitance $C$ of the capacitor.
Examiner tip
Marks: (1) large triangle / correct coordinates; (2) gradient $= 12$ ; (3) correct units ( $\mu\text{C V}^{-1}$ , i.e. $\mu\text{F}$ ) AND identifying it as capacitance. Dropping the units is the single most common lost mark; a small triangle loses precision and the accuracy mark.
4Linearising capacitor discharge to find RC (4 marks)
Building confidenceShow that / prove• Edexcel IAL Physics Unit 6 (WPH16), Unit 4 link• linearisation, ln graph, capacitor, Unit 6
Question
A capacitor discharges through a resistor so that its charge follows $Q = Q_{0}e^{-t/RC}$ . Readings of $Q$ (in mC) at $t = 0,\ 20,\ 40,\ 60$ s are $50.0,\ 30.3,\ 18.4,\ 11.2$ mC. State what to plot to get a straight line, and use the graph to find the time constant $RC$ . (4 marks)
Step-by-step solution
1. Step 1
  Take natural logs of both sides (the base is $e$ , so use ln, not lg).
  $\ln Q = \ln Q_{0} - \frac{1}{RC}\,t$
2. Step 2
  Compare with $y = mx + c$ : plot $\ln Q$ (y-axis) against $t$ (x-axis). The gradient is $-\dfrac{1}{RC}$ and the intercept is $\ln Q_{0}$ .
  $\underbrace{\ln Q}_{y} = \underbrace{-\tfrac{1}{RC}}_{m}\,\underbrace{t}_{x} + \underbrace{\ln Q_{0}}_{c}$
3. Step 3
  Compute $\ln Q$ (to 3 s.f.): $\ln 50.0 = 3.91$ , $\ln 30.3 = 3.41$ , $\ln 18.4 = 2.91$ , $\ln 11.2 = 2.42$ . These fall on a straight line (the differences are equal, $\approx -0.50$ per 20 s).
4. Step 4
  Gradient from the first and last points: $\dfrac{2.42 - 3.91}{60 - 0} = -0.0248\ \text{s}^{-1}$ . Since the gradient is $-\dfrac{1}{RC}$ , the time constant is $RC = -\dfrac{1}{\text{gradient}} = \dfrac{1}{0.0248}$ .
  $RC = -\frac{1}{\text{gradient}} = \frac{1}{0.0248} = 40.3\ \text{s} \approx 40\ \text{s}$
Answer
Plot $\ln Q$ (y-axis) against $t$ (x-axis); the gradient is $-\dfrac{1}{RC}$ . Here gradient $= -0.0248\ \text{s}^{-1}$ , so $RC = 40\ \text{s}$ (intercept $= \ln Q_{0} = 3.91$ ).
Examiner tip
Marks: (1) take ln of both sides; (2) axes $\ln Q$ vs $t$ and gradient $= -1/RC$ ; (3) $\ln Q$ values to 3 s.f.; (4) $RC \approx 40\ \text{s}$ . Use ln (base $e$ ), not lg — mixing them scales the gradient by $\ln 10 \approx 2.303$ . The same method with $\ln V$ or $\ln I$ works because $V$ and $I$ share the same exponential.
5ln A vs t to find the decay constant λ (4 marks)
Building confidenceShow that / prove• Edexcel IAL Physics Unit 6 (WPH16), Unit 5 link• linearisation, ln graph, radioactive decay, Unit 6
Question
The corrected count rate $A$ of a radioactive source follows $A = A_{0}e^{-\lambda t}$ . Readings of $A$ at $t = 0,\ 5,\ 10,\ 15$ min are $480,\ 302,\ 190,\ 120$ counts s⁻¹. State what to plot to obtain a straight line, and determine the decay constant $\lambda$ from the graph. (4 marks)
Step-by-step solution
1. Step 1
  Take natural logs of both sides of the exponential decay law.
  $\ln A = \ln A_{0} - \lambda t$
2. Step 2
  Compare with $y = mx + c$ : plot $\ln A$ (y-axis) against $t$ (x-axis). The gradient is $-\lambda$ and the intercept is $\ln A_{0}$ .
  $\underbrace{\ln A}_{y} = \underbrace{-\lambda}_{m}\,\underbrace{t}_{x} + \underbrace{\ln A_{0}}_{c}$
3. Step 3
  Compute $\ln A$ (to 3 s.f.): $\ln 480 = 6.17$ , $\ln 302 = 5.71$ , $\ln 190 = 5.25$ , $\ln 120 = 4.79$ . These fall on a straight line (steps of $\approx -0.46$ per 5 min).
4. Step 4
  Gradient from the first and last points: $\dfrac{4.79 - 6.17}{15 - 0} = -0.0920\ \text{min}^{-1}$ . The gradient is $-\lambda$ , so $\lambda = 0.0920\ \text{min}^{-1}$ ( $= 1.53\times10^{-3}\ \text{s}^{-1}$ ).
  $\lambda = -\text{gradient} = 0.0920\ \text{min}^{-1} = 1.53\times10^{-3}\ \text{s}^{-1}$
Answer
Plot $\ln A$ (y-axis) against $t$ (x-axis); gradient $= -\lambda$ , so $\lambda = 0.0920\ \text{min}^{-1} = 1.53\times10^{-3}\ \text{s}^{-1}$ (intercept $= \ln A_{0} = 6.17$ ).
Examiner tip
Marks: (1) take ln of both sides; (2) axes $\ln A$ vs $t$ , gradient $= -\lambda$ ; (3) $\ln A$ values to 3 s.f.; (4) $\lambda = 0.0920\ \text{min}^{-1}$ . Two classic errors: forgetting the minus sign (so $\lambda$ comes out negative), and using raw count $A$ that has not had the background subtracted first — always correct for background before taking logs.
6Log–log plot to confirm an inverse-square law (4 marks)
Building confidenceShow that / prove• Edexcel IAL Physics Unit 6 (WPH16), Unit 5 link• linearisation, log graph, inverse square, Unit 6
Question
The intensity $I$ of gamma radiation is thought to obey a power law with distance $r$ : $I = k\,r^{n}$ . A student wants to confirm the inverse-square law using a log–log graph. State what to plot, explain how the gradient tests the law, and state the value the gradient should take if $I \propto 1/r^{2}$ . (4 marks)
Step-by-step solution
1. Step 1
  Take logarithms (base 10) of both sides to turn the power into a multiplier and the product into a sum.
  $\lg I = \lg(k\,r^{n}) = \lg k + n\lg r$
2. Step 2
  Compare with $y = mx + c$ : plot $\lg I$ (y-axis) against $\lg r$ (x-axis). The gradient is $n$ (the power) and the intercept is $\lg k$ .
  $\underbrace{\lg I}_{y} = \underbrace{n}_{m}\,\underbrace{\lg r}_{x} + \underbrace{\lg k}_{c}$
3. Step 3
  A straight line confirms that $I$ is a power of $r$ . The gradient equals the power $n$ , which the student reads directly from the graph.
4. Step 4
  For an inverse-square law $I \propto r^{-2}$ , so $n = -2$ . Thus a gradient of $\mathbf{-2}$ (within uncertainty) confirms the inverse-square law.
  $n = \text{gradient} = -2 \;\Rightarrow\; I \propto \frac{1}{r^{2}}$
Answer
Plot $\lg I$ (y-axis) against $\lg r$ (x-axis). A straight line confirms a power law; the gradient equals the power $n$ . If the gradient is $-2$ (within uncertainty), the inverse-square law $I \propto 1/r^{2}$ is confirmed.
Examiner tip
Marks: (1) take logs of both sides; (2) plot $\lg I$ vs $\lg r$ ; (3) gradient $= n$ / straight line confirms power law; (4) gradient $= -2$ . Use log–log (any base) for a power law and ln vs $x$ for an exponential — do not confuse the two. An alternative direct test is to plot $I$ against $1/r^{2}$ , which should give a straight line through the origin.
7Combining uncertainties with a squared term (4 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 6 (WPH16), Unit 4 link• percentage uncertainty, combining, SHM, Unit 6
Question
A mass–spring oscillator is used to find the spring constant from $k = \dfrac{4\pi^{2}m}{T^{2}}$ . The mass is $m = 0.250 \pm 0.001\ \text{kg}$ and the period is $T = 0.960 \pm 0.008\ \text{s}$ . Calculate $k$ and its percentage uncertainty. (4 marks)
Step-by-step solution
1. Step 1
  Calculate $k$ from the mean values.
  $k = \frac{4\pi^{2}(0.250)}{(0.960)^{2}} = \frac{9.8696}{0.9216} = 10.7\ \text{N m}^{-1}$
2. Step 2
  Find the percentage uncertainty in each measured quantity (absolute ÷ value × 100).
  $\frac{0.001}{0.250}\times100 = 0.40\%, \qquad \frac{0.008}{0.960}\times100 = 0.833\%$
3. Step 3
  $k = 4\pi^{2}m/T^{2}$ is a product/quotient, so add the percentage uncertainties — but $T$ is squared, so its percentage uncertainty is multiplied by the power 2. (The constant $4\pi^{2}$ has no uncertainty.)
  $\%\,k = \%m + 2\times\%T = 0.40 + 2(0.833)$
4. Step 4
  Evaluate the total percentage uncertainty.
  $\%\,k = 0.40 + 1.67 = 2.07\% \approx 2.1\%$
Answer
$k = 10.7\ \text{N m}^{-1}$ with a percentage uncertainty of about $2.1\%$ (i.e. $10.7 \pm 0.2\ \text{N m}^{-1}$ ).
Examiner tip
Marks: (1) $k = 10.7\ \text{N m}^{-1}$ ; (2) both %uncertainties; (3) recognising the factor of 2 for the squared $T$ ; (4) total $\approx 2.1\%$ . The dominant error is forgetting to double the period's contribution — a squared term contributes twice its percentage uncertainty. $T$ dominates, so timing more oscillations to cut $\%T$ is the best improvement.
8Counting statistics uncertainty of a radioactive count (4 marks)
StretchMulti-step problem• Edexcel IAL Physics Unit 6 (WPH16), Unit 5 link• counting uncertainty, percentage uncertainty, radioactivity, Unit 6
Question
A Geiger counter records $N = 400$ counts in a fixed time. Background over the same time is negligible. (a) State the absolute and percentage uncertainty in the count. (b) The student wants the percentage uncertainty reduced to 2%. Determine how many counts must be recorded. (4 marks)
Step-by-step solution
1. Step 1
  For a radioactive count, the readings follow counting statistics: the absolute uncertainty in a count of $N$ is $\sqrt{N}$ .
  $\Delta N = \sqrt{N} = \sqrt{400} = 20$
2. Step 2
  The percentage uncertainty is $\dfrac{\sqrt{N}}{N}\times100 = \dfrac{1}{\sqrt{N}}\times100$ .
  $\%\,N = \frac{\sqrt{400}}{400}\times100 = \frac{20}{400}\times100 = 5.0\%$
3. Step 3
  For a target of $2\%$ , set $\dfrac{1}{\sqrt{N}}\times100 = 2$ , so $\sqrt{N} = \dfrac{100}{2} = 50$ .
  $\frac{100}{\sqrt{N}} = 2 \;\Rightarrow\; \sqrt{N} = 50$
4. Step 4
  Square to find the required count. Recording more counts (a longer count time) shrinks the percentage uncertainty.
  $N = 50^{2} = 2500 \text{ counts}$
Answer
(a) $\Delta N = \sqrt{400} = 20$ , so the percentage uncertainty is $5.0\%$ . (b) To reach $2\%$ the student must record $N = 2500$ counts.
Examiner tip
Marks: (1) $\Delta N = \sqrt{N} = 20$ ; (2) $\%N = 5.0\%$ ; (3) set $100/\sqrt{N} = 2$ ; (4) $N = 2500$ . The key IA2 idea: for counting statistics the uncertainty is $\sqrt{N}$ , so the percentage uncertainty falls as $1/\sqrt{N}$ — quadrupling the count halves the % uncertainty. Forgetting to subtract background before this analysis is a common evaluation slip.
9Uncertainty in a gradient by the max/min line method (5 marks)
StretchGraph or diagram• Edexcel IAL Physics Unit 6 (WPH16) style• gradient uncertainty, max/min line, ln plot, Unit 6
Question
On a $\ln A$ vs $t$ graph for a decay, a student draws a best-fit line of gradient $-0.0920\ \text{min}^{-1}$ , a steepest reasonable line of gradient $-0.0980\ \text{min}^{-1}$ and a shallowest line of gradient $-0.0860\ \text{min}^{-1}$ , both still passing through the error bars. Determine the decay constant $\lambda$ and the percentage uncertainty in the gradient. (5 marks)
Step-by-step solution
1. Step 1
  The gradient is $-\lambda$ , so the best-fit line gives $\lambda = 0.0920\ \text{min}^{-1}$ . The max and min lines are the steepest and shallowest that still pass through the error bars; their gradients bracket the best-fit gradient.
  $m_{\text{best}} = -0.0920, \quad m_{\text{steep}} = -0.0980, \quad m_{\text{shallow}} = -0.0860\ \text{min}^{-1}$
2. Step 2
  The absolute uncertainty in the gradient is half the range of the steepest and shallowest gradients (work with magnitudes).
  $\Delta m = \frac{|m_{\text{steep}}| - |m_{\text{shallow}}|}{2} = \frac{0.0980 - 0.0860}{2}$
3. Step 3
  Evaluate the absolute uncertainty.
  $\Delta m = \frac{0.0120}{2} = 0.0060\ \text{min}^{-1}$
4. Step 4
  The percentage uncertainty is the absolute uncertainty divided by the magnitude of the best-fit gradient.
  $\%\,m = \frac{\Delta m}{|m_{\text{best}}|}\times100 = \frac{0.0060}{0.0920}\times100$
5. Step 5
  Evaluate. Since $\lambda = -\text{gradient}$ (a single power of the gradient), the percentage uncertainty in $\lambda$ equals this.
  $\%\,m = 6.5\% \;\Rightarrow\; \lambda = 0.0920 \pm 6.5\%$
Answer
$\lambda = 0.0920\ \text{min}^{-1}$ ; the gradient (and hence $\lambda$ ) has a percentage uncertainty of about $6.5\%$ ( $\Delta m = 0.0060\ \text{min}^{-1}$ ).
Examiner tip
Marks: (1) $\lambda = 0.0920$ and max/min lines defined via error bars; (2) $\Delta m = (\text{max}-\text{min})/2$ using magnitudes; (3) $\Delta m = 0.0060$ ; (4) %method; (5) $6.5\%$ . Common slips: taking the whole range instead of half, dividing by the max gradient rather than the best-fit gradient, and getting confused by the negative gradients — use magnitudes throughout. Above ~5% signals scatter worth reducing (e.g. longer count times).

Model Answers — Practical Skills II - Processing Results

High-scoring sample answers for practical skills ii - processing results 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). State what is meant by ‘linearising’ a relationship, and explain why it is useful when processing results.
Model answer
Linearisation means rearranging a non-linear relationship (such as an exponential or a power law) into the straight-line form $y = mx + c$ , so that plotting the two rearranged quantities gives a straight line. For example, $A = A_{0}e^{-\lambda t}$ becomes $\ln A = \ln A_{0} - \lambda t$ , which is linear in $t$ .

It is useful because a straight line is easy to test and to read: its gradient and intercept give the physical constants directly (here $-\lambda$ and $\ln A_{0}$ ), a best-fit line averages out random scatter, and any curvature would immediately reveal that the assumed relationship is wrong.
Why this scores
Why this scores 2/2. (1) rearranging a non-linear law into $y = mx + c$ form / a straight line; (2) a reason — gradient and intercept give constants, or a straight line is easy to test and averages scatter. A worked example (ln A vs t) strengthens the answer but is not required.
Question 2
2 marks
Q (2 marks). A student has data that may follow either an exponential law $y = Ae^{kx}$ or a power law $y = Ax^{n}$ . State which log they should use for each case and what they should plot.
Model answer
For an exponential $y = Ae^{kx}$ , take natural logs (ln): $\ln y = kx + \ln A$ . Plot $\ln y$ against $x$ (a lin–log plot); a straight line confirms the exponential and the gradient is $k$ .

For a power law $y = Ax^{n}$ , take logs to base 10 (lg): $\lg y = n\lg x + \lg A$ . Plot $\lg y$ against $\lg x$ (a log–log plot); a straight line confirms the power law and the gradient is the power $n$ .
Why this scores
Why this scores 2/2. (1) exponential → ln, plot $\ln y$ vs $x$ ; (2) power law → lg (log–log), plot $\lg y$ vs $\lg x$ . Mixing them (e.g. plotting $\lg y$ vs $x$ for an exponential) is the classic error and only scales the gradient by $\ln 10$ ; the marker credits the correct axes and log base for each case.
Question 3
3 marks
Q (3 marks). A Geiger–Müller tube records 900 counts from a source in a fixed time (background negligible). State the absolute uncertainty in this count, calculate the percentage uncertainty, and state one way to reduce it.
Model answer
For a radioactive count, counting statistics give an absolute uncertainty of $\Delta N = \sqrt{N}$ :

$\Delta N = \sqrt{900} = 30 \text{ counts}$

The percentage uncertainty is:

$\%N = \frac{\sqrt{900}}{900}\times100 = \frac{30}{900}\times100 = 3.3\%$

To reduce it, record more counts — count for a longer time (or use a more active source). Because the percentage uncertainty is $100/\sqrt{N}$ , it falls as $N$ increases; e.g. quadrupling the counts halves the percentage uncertainty.
Why this scores
Why this scores 3/3. (1) $\Delta N = \sqrt{N} = 30$ ; (2) $\%N = 3.3\%$ ; (3) a valid reduction — count for longer / more counts, justified by the $1/\sqrt{N}$ dependence. Quoting $\Delta N = N$ or a fixed % regardless of $N$ loses marks.
Question 4
Edexcel IAL Physics Unit 6 (WPH16), Unit 4 style4 marks
Q (4 marks). A capacitor discharges through a resistor and a student plots $\ln V$ (y-axis) against $t$ (x-axis), obtaining a straight line of gradient $-0.040\ \text{s}^{-1}$ . The resistor used is $R = 5.0\times10^{5}\ \Omega$ . (a) State why plotting $\ln V$ against $t$ gives a straight line. (b) Determine the capacitance $C$ .
Model answer
(a) The discharge follows $V = V_{0}e^{-t/RC}$ . Taking natural logs of both sides:

$\ln V = \ln V_{0} - \frac{1}{RC}\,t$

Comparing with $y = mx + c$ , a graph of $\ln V$ (y) against $t$ (x) is a straight line with gradient $= -\dfrac{1}{RC}$ and intercept $\ln V_{0}$ .

(b) The time constant is $RC = -\dfrac{1}{\text{gradient}} = \dfrac{1}{0.040} = 25\ \text{s}$ . Then:

$C = \frac{RC}{R} = \frac{25}{5.0\times10^{5}} = 5.0\times10^{-5}\ \text{F} = 50\ \mu\text{F}$
Why this scores
Why this scores 4/4. (a) 2 marks: take ln, so $\ln V = \ln V_{0} - t/RC$ , gradient $= -1/RC$ . (b) 2 marks: $RC = 1/0.040 = 25\ \text{s}$ , then $C = RC/R = 50\ \mu\text{F}$ . Forgetting the minus sign (so $RC$ comes out negative) or using lg instead of ln are the usual errors.
Question 5
Edexcel IAL Physics Unit 6 (WPH16) style4 marks
Q (4 marks). A quantity is calculated from $P = \dfrac{a b^{2}}{c}$ where $a = 2.00 \pm 0.02$ , $b = 5.00 \pm 0.10$ and $c = 4.00 \pm 0.04$ (all in consistent units). Calculate $P$ and its percentage uncertainty.
Model answer
The value is:

$P = \frac{(2.00)(5.00)^{2}}{4.00} = \frac{2.00 \times 25.0}{4.00} = 12.5$

The percentage uncertainties are:

$\%a = \frac{0.02}{2.00}\times100 = 1.0\%, \quad \%b = \frac{0.10}{5.00}\times100 = 2.0\%, \quad \%c = \frac{0.04}{4.00}\times100 = 1.0\%$

Because $b$ is squared, its percentage uncertainty is doubled; $a$ and $c$ contribute once each (product/quotient):

$\%P = \%a + 2\times\%b + \%c = 1.0 + 2(2.0) + 1.0 = 6.0\%$

So $P = 12.5$ with a percentage uncertainty of about 6.0%.
Why this scores
Why this scores 4/4. (1) $P = 12.5$ ; (2) the three %uncertainties; (3) recognising the factor of 2 on $\%b$ ; (4) total $= 6.0\%$ . The standard error is adding $\%b$ once (giving 4%) instead of doubling it — a squared term contributes TWICE its percentage uncertainty.
Question 6
Edexcel IAL Physics Unit 6 (WPH16), Unit 5 style4 marks
Q (4 marks). The intensity $I$ of gamma radiation from a point source is measured at several distances $r$ and is thought to obey an inverse-square law. A student plots $\lg I$ against $\lg r$ . State what a straight line confirms, what the gradient should be if the inverse-square law holds, and one alternative graph that would test the same law.
Model answer
Writing the relationship as a power law $I = k r^{n}$ and taking logs (base 10):

$\lg I = n\lg r + \lg k$

A straight line on the $\lg I$ – $\lg r$ plot confirms that $I$ is a power of $r$ (i.e. the power-law form holds), with the gradient equal to the power $n$ .

If the inverse-square law $I \propto r^{-2}$ holds, the gradient should equal $-2$ (within experimental uncertainty).

An alternative test is to plot $I$ against $1/r^{2}$ : if the inverse-square law holds this is a straight line through the origin, with gradient equal to the constant $k$ .
Why this scores
Why this scores 4/4. (1) straight line confirms a power law; (2) gradient $= n$ ; (3) gradient $= -2$ for inverse square; (4) alternative graph $I$ vs $1/r^{2}$ (straight line through origin). Background must be subtracted from $I$ first — a good answer mentions this.
Question 7
Edexcel IAL Physics Unit 6 (WPH16) style4 marks
Q (4 marks). In an evaluation of a radioactive-decay experiment, a student writes that ‘taking more counts makes the experiment more valid’. Explain the difference between the validity and the reliability of an experiment, and state correctly what taking more counts improves.
Model answer
Validity concerns whether the experiment actually measures what it is intended to measure — a fair test with systematic errors removed. In a decay experiment, validity requires the background count to be subtracted; if it is not, the result is invalid however many counts are taken.

Reliability concerns whether the results are consistent and repeatable — small random scatter when the measurement is repeated.

The student is incorrect: taking more counts reduces the counting (random) uncertainty ( $\%N = 100/\sqrt{N}$ falls as $N$ rises), so it improves the reliability/precision, not the validity. Validity is improved by removing systematic error (e.g. subtracting background, correcting for dead time), not by taking more counts.
Why this scores
Why this scores 4/4. (1) validity = measures what is intended / fair test with systematic errors removed; (2) reliability = consistent, repeatable results; (3) more counts reduces random/counting error → improves reliability; (4) correcting the student — it does NOT improve validity. The validity/reliability mix-up is one of the most common evaluation errors examiners flag.
Question 8
Edexcel IAL Physics Unit 6 (WPH16), Unit 5 style6 marks
Q (6 marks). A student measures the count rate of a gamma source at various distances to test the inverse-square law, timing each count for 10 s with the background not subtracted, and obtains a log–log gradient of $-1.6$ rather than the expected $-2$ . (a) Identify the main systematic error and explain how it distorts the gradient. (b) Describe two specific improvements to the method, explaining how each helps.
Model answer
(a) The main systematic error is failing to subtract the background count rate. Background adds a roughly constant amount to every reading. At large $r$ the source's own count is small, so the constant background is a larger fraction of the reading there, lifting the far points up. This flattens the fall-off, making the log–log line less steep — pulling the gradient from $-2$ towards a smaller magnitude (here $-1.6$ ).

(b) Two specific improvements:

Measure the background count rate separately (source removed) and subtract it from every reading before taking logs. This removes the constant offset, so the corrected intensities follow the true inverse-square fall-off and the gradient returns towards $-2$ .

Count for much longer at each distance (e.g. 100 s or more), especially at large $r$ . A longer count raises $N$ , so the counting uncertainty $\%N = 100/\sqrt{N}$ falls; the far points, where the count is smallest, become far more reliable and the best-fit gradient is better defined. Repeating and averaging further reduces random scatter.
Why this scores
Why this scores 6/6. (a) 2 marks: background not subtracted AND the mechanism — constant offset is a bigger fraction at large $r$ , flattening the gradient. (b) 4 marks: two SPECIFIC, feasible improvements EACH with the mechanism — subtract background (removes the offset), and count longer (raises $N$ , lowering counting uncertainty at large $r$ ). Vague answers (‘be more accurate’) score nothing; improvements must target the named error.
Question 9
Edexcel IAL Physics Unit 6 (WPH16), Unit 5 style6 marks
Q (6 marks). A student plots $\ln A$ (y-axis) against $t$ (x-axis) for a decaying source and obtains a straight line, except for one point that lies clearly off the line. (a) State what this off-line point is called and how the student should treat it when drawing the best-fit line. (b) The best-fit line passes through $(0,\ 5.30)$ and $(200\ \text{s},\ 3.30)$ . Determine the gradient with units and hence the decay constant $\lambda$ .
Model answer
(a) The off-line point is an anomaly (anomalous result). The student should ignore it when drawing the best-fit line (draw the line through the trend of the remaining points) and, if possible, check it by repeating that reading; it must not be allowed to pull the line towards it. A likely cause here is a fluctuation in the count or a mis-subtracted background.

(b) Using a large triangle between the two given points:

$\text{gradient} = \frac{\Delta(\ln A)}{\Delta t} = \frac{3.30 - 5.30}{200 - 0} = \frac{-2.00}{200} = -0.0100\ \text{s}^{-1}$

For a $\ln A$ – $t$ graph the gradient is $-\lambda$ , so:

$\lambda = -\text{gradient} = 0.0100\ \text{s}^{-1}$
Why this scores
Why this scores 6/6. (a) 3 marks: named as an anomaly; ignore it for the best-fit line / do not let it pull the line; check by repeating. (b) 3 marks: gradient $= -0.0100\ \text{s}^{-1}$ with units from a large triangle; gradient $= -\lambda$ ; $\lambda = 0.0100\ \text{s}^{-1}$ . Forgetting the minus sign (quoting $\lambda$ negative) or omitting the gradient's units are the usual lost marks.
Question 10
Edexcel IAL Physics Unit 6 (WPH16), Unit 4 style6 marks
Q (6 marks). A student investigates how the period $T$ of a mass on a spring depends on the mass $m$ , expecting $T = 2\pi\sqrt{\dfrac{m}{k}}$ . (a) Show that plotting $T^{2}$ against $m$ should give a straight line through the origin, and state what the gradient represents. (b) The best-fit line has gradient $2.40\ \text{s}^{2}\,\text{kg}^{-1}$ , with a steepest line of $2.58$ and a shallowest line of $2.22\ \text{s}^{2}\,\text{kg}^{-1}$ . Determine the spring constant $k$ and the percentage uncertainty in the gradient.
Model answer
(a) Squaring the relationship:

$T^{2} = \frac{4\pi^{2}}{k}\,m$

Comparing with $y = mx$ (with $y = T^{2}$ , $x = m$ ), a graph of $T^{2}$ against $m$ is a straight line through the origin with gradient $= \dfrac{4\pi^{2}}{k}$ .

(b) Rearranging for $k$ using the best-fit gradient:

$k = \frac{4\pi^{2}}{\text{gradient}} = \frac{4\pi^{2}}{2.40} = 16.4\ \text{N m}^{-1}$

The uncertainty in the gradient (max/min method):

$\Delta m = \frac{2.58 - 2.22}{2} = 0.18\ \text{s}^{2}\,\text{kg}^{-1}$

$\%\,\text{gradient} = \frac{0.18}{2.40}\times100 = 7.5\%$

So $k = 16.4\ \text{N m}^{-1}$ and the gradient (and hence $k$ ) has a percentage uncertainty of about 7.5%.
Why this scores
Why this scores 6/6. (a) 3 marks: square to $T^{2} = (4\pi^{2}/k)m$ , so $T^{2}\propto m$ , gradient $= 4\pi^{2}/k$ . (b) 3 marks: $k = 4\pi^{2}/\text{gradient} = 16.4\ \text{N m}^{-1}$ ; $\Delta m = (2.58-2.22)/2 = 0.18$ ; $\%\text{gradient} = 7.5\%$ . Because $k = 4\pi^{2}/\text{gradient}$ (a single power of the gradient), the % uncertainty in $k$ equals that in the gradient. Taking the full max−min range, not half, is the classic slip.

Key Formulae — Practical Skills II - Processing Results

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

Absolute uncertainty of a single reading
$\Delta x = \tfrac{1}{2}\times\text{resolution}$
$\Delta x$
absolute uncertainty in the reading
$resolution$
smallest division / step of the instrument
When to use
For a single reading on an analogue or digital scale (e.g. a metre rule to 1 mm gives $\Delta x = 0.5\ \text{mm}$ ).
Example
Micrometer resolution $0.01\ \text{mm}$ : single reading uncertainty $= 0.005\ \text{mm}$ .
Absolute uncertainty from repeated readings
$\Delta x = \frac{x_{\text{max}} - x_{\text{min}}}{2}$
$x_{\text{max}}$
largest repeated value
$x_{\text{min}}$
smallest repeated value
$\Delta x$
absolute uncertainty (half the spread)
When to use
When several repeats of the same measurement are taken; the mean is quoted with this ± uncertainty (half the range).
Example
Times 20.1, 20.3, 20.5 s: mean 20.3 s, $\Delta t = (20.5-20.1)/2 = 0.2\ \text{s}$ .
Percentage uncertainty
$\%\,x = \frac{\Delta x}{x}\times100\%$
$\Delta x$
absolute uncertainty
$x$
measured value
When to use
Converting an absolute uncertainty into a fraction of the reading, so uncertainties in different quantities can be combined.
Example
$R = 3.4 \pm 0.1\ \Omega$ : $\%R = (0.1/3.4)\times100 = 2.9\%$ .
Combining uncertainties (products and quotients)
$\text{if } Q = \frac{a\,b}{c} \;\Rightarrow\; \%Q = \%a + \%b + \%c$
$Q$
derived quantity
$\%a,\ \%b,\ \%c$
percentage uncertainties of the measured quantities
When to use
For any quantity formed by multiplying and/or dividing measurements: ADD the percentage uncertainties (never the absolute ones).
Example
$\rho = RA/L$ : $\%\rho = \%R + \%A + \%L$ .
Combining uncertainties (powers)
$\text{if } Q = x^{n} \;\Rightarrow\; \%Q = |n|\times\%x$
$n$
power to which the quantity is raised
$\%x$
percentage uncertainty in $x$
When to use
When a measured quantity is raised to a power. A squared term ( $n=2$ ) contributes TWICE its percentage uncertainty; in $k = 4\pi^{2}m/T^{2}$ the $T^{2}$ term contributes $2\,\%T$ .
Example
$T$ has $\%T = 0.833\%$ ; in $k\propto 1/T^{2}$ it contributes $2\times0.833 = 1.67\%$ .
Counting (Poisson) uncertainty of a radioactive count
$\Delta N = \sqrt{N}, \qquad \%\,N = \frac{\sqrt{N}}{N}\times100 = \frac{100}{\sqrt{N}}$
$N$
number of counts recorded
$\Delta N$
absolute uncertainty in the count ( $=\sqrt{N}$ )
When to use
For a single radioactive count $N$ (counting statistics). The percentage uncertainty falls as $1/\sqrt{N}$ , so recording more counts (longer count time) improves precision. Subtract background first.
Example
$N = 400$ : $\Delta N = 20$ , $\%N = 5.0\%$ . To reach $2\%$ needs $N = (100/2)^{2} = 2500$ .
Gradient of a straight-line graph
$m = \frac{\Delta y}{\Delta x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m$
gradient (units = y-units ÷ x-units)
$(x_{1},y_{1}),(x_{2},y_{2})$
two points on the best-fit line, far apart
When to use
Reading a rate/constant from a linear graph. Use a LARGE triangle (≥ half the line) and always STATE the units of the gradient.
Example
$Q$ – $V$ line through $(0,0)$ and $(8.0,96)$ : $m = 96/8.0 = 12\ \mu\text{F} = C$ .
Uncertainty in a gradient (max/min lines)
$\Delta m = \frac{m_{\text{max}} - m_{\text{min}}}{2}$
$m_{\text{max}}$
gradient of the steepest reasonable line
$m_{\text{min}}$
gradient of the shallowest reasonable line
$\Delta m$
absolute uncertainty in the gradient
When to use
Estimating the uncertainty in a best-fit gradient from data with error bars; the max/min lines still pass through the error bars. For negative gradients, use magnitudes.
Example
$|m_{\text{steep}}|=0.0980$ , $|m_{\text{shallow}}|=0.0860$ : $\Delta m = 0.0120/2 = 0.0060$ .
Linearising an exponential (ln plot)
$y = y_{0}e^{-kx} \;\Rightarrow\; \ln y = \ln y_{0} - kx$
$k$
decay constant; gradient of $\ln y$ vs $x$ is $-k$
$y_{0}$
initial value; $\ln y_{0}$ = y-intercept
When to use
Testing/using an exponential relationship. Plot $\ln y$ (base $e$ , NOT lg) against $x$ ; the gradient is $-k$ (negative for decay), the intercept is $\ln y_{0}$ .
Example
Capacitor $Q = Q_{0}e^{-t/RC}$ : plot $\ln Q$ vs $t$ , gradient $= -1/RC$ . Decay $A = A_{0}e^{-\lambda t}$ : gradient $= -\lambda$ . Absorption $I = I_{0}e^{-\mu x}$ : gradient $= -\mu$ .
Linearising a power law (log–log plot)
$y = Ax^{n} \;\Rightarrow\; \lg y = n\lg x + \lg A$
$n$
power = gradient of the $\lg y$ vs $\lg x$ line
$A$
constant; $\lg A$ = y-intercept, so $A = 10^{\text{intercept}}$
When to use
Testing/using a power-law relationship, including confirming an inverse-square law. Plot $\lg y$ against $\lg x$ ; the gradient gives the power $n$ (e.g. $n = -2$ confirms inverse square).
Example
$I = k r^{n}$ : plot $\lg I$ vs $\lg r$ ; a gradient of $-2$ confirms $I \propto 1/r^{2}$ .
Inverse-square law (direct linear test)
$I \propto \frac{1}{r^{2}} \;\Rightarrow\; \text{plot } I \text{ against } \frac{1}{r^{2}}$
$I$
intensity (y-axis)
$1/r^{2}$
reciprocal square of distance (x-axis)
When to use
A direct alternative to a log–log plot: plotting $I$ against $1/r^{2}$ gives a straight line through the origin if the inverse-square law holds. The gradient is the constant $k$ .
Example
Gamma from a point source: $I$ vs $1/r^{2}$ is a straight line through the origin.

Key Definitions and Keywords — Practical Skills II - Processing Results

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

Line of best fit
Examiner keyword
A single smooth line (straight or curved) drawn through plotted points so they are balanced about it, representing the trend and averaging out random scatter. It is NOT a dot-to-dot line.
Gradient
Examiner keyword
The slope of a graph, $m = \dfrac{\Delta y}{\Delta x}$ , read from a large triangle spanning at least half the line. Its units are the y-axis units divided by the x-axis units, and it often equals a physical quantity (e.g. $-\lambda$ on a $\ln A$ – $t$ plot).
y-intercept
Examiner keyword
The value of $y$ where the best-fit line crosses the y-axis ( $x = 0$ ). In $y = mx + c$ it is the constant $c$ , and it often has a physical meaning (e.g. $\ln Q_{0}$ or $\lg k$ on a log graph).
Linearisation
Examiner keyword
Rearranging a non-linear relationship into the straight-line form $y = mx + c$ so a graph gives a constant from its gradient/intercept — e.g. $\ln A = \ln A_{0} - \lambda t$ (plot $\ln A$ vs $t$ ) or $\lg y = n\lg x + \lg A$ (plot $\lg y$ vs $\lg x$ ).
ln (lin–log) plot
Examiner keyword
A plot of $\ln y$ against $x$ , used for an exponential $y = y_{0}e^{-kx}$ . A straight line confirms the exponential; the gradient is $-k$ (e.g. $-\lambda$ , $-1/RC$ or $-\mu$ ) and the intercept is $\ln y_{0}$ . Uses natural logs (base $e$ ).
log–log plot
Examiner keyword
A plot of $\lg y$ against $\lg x$ , used for a power law $y = Ax^{n}$ . A straight line confirms the power law; the gradient equals the power $n$ (e.g. $-2$ for an inverse-square law) and the intercept is $\lg A$ .
Decay constant (λ)
Examiner keyword
The probability per unit time that a given nucleus decays; the constant in $A = A_{0}e^{-\lambda t}$ . It equals minus the gradient of a $\ln A$ vs $t$ graph and has units of $\text{s}^{-1}$ .
Time constant (RC)
The product $RC$ for a discharging capacitor, in seconds; the time for the charge to fall to $1/e$ of its value. It equals minus the reciprocal of the gradient of a $\ln Q$ (or $\ln V$ ) vs $t$ graph.
Absolute uncertainty
Examiner keyword
The size of the possible error in a reading, in the same units as the measurement: $\tfrac{1}{2}\times$ resolution for a single reading, or $\dfrac{x_{\text{max}}-x_{\text{min}}}{2}$ for repeats. Quoted as value ± uncertainty.
Percentage uncertainty
Examiner keyword
The absolute uncertainty expressed as a percentage of the reading, $\dfrac{\Delta x}{x}\times100\%$ . Percentage uncertainties are added for products/quotients and multiplied by the power for a raised term.
Counting uncertainty
Examiner keyword
The uncertainty in a radioactive count $N$ from counting statistics: absolute $\Delta N = \sqrt{N}$ , so percentage $= 100/\sqrt{N}$ . Recording more counts (longer time) reduces the percentage uncertainty (which falls as $1/\sqrt{N}$ ).
Systematic error
Examiner keyword
An error that shifts all readings by the same amount in the same direction (e.g. a zero error, un-subtracted background, or voltmeter loading). It affects accuracy and is NOT reduced by repeating.
Random error
Examiner keyword
An error causing readings to scatter unpredictably about the true value (e.g. count fluctuations, reaction time). It affects precision/reliability and is reduced by repeating and averaging (or, for counts, recording more counts).
Validity
Examiner keyword
Whether an experiment actually measures what it is intended to measure — a fair test with the other variables controlled and systematic errors removed (e.g. background subtracted). Repeating readings does NOT improve validity.
Reliability
Examiner keyword
Whether results are consistent and repeatable — small random scatter when the measurement is repeated. Improved by repeating readings and taking a mean (or, for counts, taking more counts).
Maximum/minimum line
The steepest and shallowest straight lines that still pass through (most of) the error bars. Their gradients bracket the best-fit gradient and give the gradient uncertainty, $\Delta m = \dfrac{m_{\text{max}}-m_{\text{min}}}{2}$ .

Common Mistakes and Misconceptions — Practical Skills II - Processing Results

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

✕Confusing ln (base e) with log (base 10) when linearising — e.g. plotting lg for an exponential
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
Both ‘log’ buttons exist on the calculator and the difference in method is not memorised.
How to avoid it
Use ln (base $e$ ) for exponentials $y = y_{0}e^{-kx}$ (plot $\ln y$ vs $x$ , gradient $= -k$ ), and lg (base 10) for power laws $y = Ax^{n}$ (plot $\lg y$ vs $\lg x$ , gradient $= n$ ). Mixing them scales the gradient by $\ln 10 \approx 2.303$ .
✕Forgetting the minus sign when reading $\lambda$ or $1/RC$ from the gradient of a $\ln$ plot
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
Students read the gradient's magnitude but overlook that the decay gradient is negative, so $\lambda = -\text{gradient}$ .
How to avoid it
Write the linearised form first: $\ln A = \ln A_{0} - \lambda t$ , so the gradient is $-\lambda$ (negative). Therefore $\lambda = -\text{gradient}$ , a positive value. The same applies to $-1/RC$ and $-\mu$ . A negative $\lambda$ or $RC$ is a sure sign of this error.
✕Not multiplying the percentage uncertainty by the power for a squared (or cubed) term
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
Students add each measurement's percentage uncertainty once, ignoring that some are raised to a power.
How to avoid it
For a term $x^{n}$ the percentage uncertainty is $|n|\times\%x$ . A squared quantity (e.g. $T^{2}$ in $k = 4\pi^{2}m/T^{2}$ , or $A\propto d^{2}$ ) contributes twice its percentage uncertainty; a cube contributes three times.
✕Plotting the raw data (e.g. $A$ vs $t$ , a curve) instead of the linearised quantities
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
Students plot what was measured directly rather than rearranging to the straight-line form first.
How to avoid it
Decide the linearisation before plotting: for an exponential plot $\ln y$ vs $x$ ; for a power law plot $\lg y$ vs $\lg x$ . A curved raw graph cannot give a gradient equal to $-\lambda$ , $-1/RC$ or $n$ — only the linearised straight line can.
✕Using a small gradient triangle (a fraction of the line) to read the gradient
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
Students pick two convenient nearby grid points instead of two well-separated points on the line.
How to avoid it
Draw a large triangle spanning at least half the plotted line and read the coordinates to the scale's precision. A big triangle reduces the fractional error in $\Delta x$ and $\Delta y$ and protects the accuracy mark.
✕Mixing up validity and reliability — e.g. saying ‘taking more counts makes the result more valid’
Edexcel IAL Physics Unit 6 (WPH16) examiner reports
Why it happens
The two evaluation terms sound similar and are used loosely in everyday speech.
How to avoid it
Reliability = consistent, repeatable results (improved by repeating/averaging or more counts, which cut random/counting error). Validity = measuring the right thing in a fair test (improved by removing systematic error, e.g. subtracting background). More counts improves reliability, NOT validity.

Practical Skills II - Processing Results — frequently asked questions

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

Practical Skills II - Processing Results

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Choosing sensible scales for a ln–linear plot (3 marks)

2Line of best fit on a linearised plot (2 marks)

3Gradient with a large triangle and its units (3 marks)

4Linearising capacitor discharge to find RC (4 marks)

5ln A vs t to find the decay constant λ (4 marks)

6Log–log plot to confirm an inverse-square law (4 marks)

7Combining uncertainties with a squared term (4 marks)

8Counting statistics uncertainty of a radioactive count (4 marks)

9Uncertainty in a gradient by the max/min line method (5 marks)

Absolute uncertainty of a single reading

Absolute uncertainty from repeated readings

Percentage uncertainty

Combining uncertainties (products and quotients)

Combining uncertainties (powers)

Counting (Poisson) uncertainty of a radioactive count

Gradient of a straight-line graph

Uncertainty in a gradient (max/min lines)

Linearising an exponential (ln plot)

Linearising a power law (log–log plot)

Inverse-square law (direct linear test)

Line of best fit

Gradient

y-intercept

Linearisation

ln (lin–log) plot

log–log plot

Decay constant (λ)

Time constant (RC)

Absolute uncertainty

Percentage uncertainty

Counting uncertainty

Systematic error

Random error

Validity

Reliability

Maximum/minimum line

✕Confusing ln (base e) with log (base 10) when linearising — e.g. plotting lg for an exponential

✕Forgetting the minus sign when reading λ\lambdaλ or 1/RC1/RC1/RC from the gradient of a ln⁡\lnln plot

✕Not multiplying the percentage uncertainty by the power for a squared (or cubed) term

✕Plotting the raw data (e.g. AAA vs ttt, a curve) instead of the linearised quantities

✕Using a small gradient triangle (a fraction of the line) to read the gradient

✕Mixing up validity and reliability — e.g. saying ‘taking more counts makes the result more valid’

Practical Skills II - Processing Results — frequently asked questions

✕Forgetting the minus sign when reading $\lambda$ or $1/RC$ from the gradient of a $\ln$ plot

✕Plotting the raw data (e.g. $A$ vs $t$ , a curve) instead of the linearised quantities