Fotoni lab visit: modern physics experiments
Why can a small glass tube, a lamp and a voltmeter tell us something about the whole universe of quantum physics?
Prior knowledge
- The stopping voltage $eU_\text{s} = hf - W$chapter Light as particles: the photoelectric effect
- A charged particle that moves across a magnetic field moves on a circlechapter Charged particles and magnetic confinement
- Light and electrons in a double slit, and probabilitychapter Waves, probability and quantum tunnelling
- Each gas gives its own line spectrumchapter From Rutherford's nucleus to Bohr's atom
- The peak of thermal radiation follows Wien's law $\lambda_\text{max} T = b$chapter When classical physics ran out of answers
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 3.
Think about it
Soon your group visits Fotoni, the open physics teaching laboratory of the University of Helsinki, on the second floor of the Physicum building on the Kumpula campus. Until now you have read about photons, electrons and spectra in this course. At Fotoni you do the experiments yourselves, with real lamps, glass tubes, coils and meters, in about an hour of lab work. Real measurements never give exactly the numbers in a book.
What can we measure about light and electrons ourselves, and how sure can we be of our numbers?
Review: the stopping voltage
Sentence starters
- A number is trustworthy if …
- We could repeat the measurement to …
- One problem could be that …
- I think the hardest part will be …, because …
Think about it: a number you can trust
Discuss with a partner: you will measure Planck's constant yourselves.
What would make your number trustworthy? Write two or three sentences. Name at least one thing that could make the result less precise.
Pair work: share your ideas aloud first, then each of you writes your own answer.
Model answer
A number is trustworthy if we can repeat the measurement and get almost the same value, and if we know how big the uncertainty of each reading is. We should also use several points and a graph instead of one reading. One problem could be that it is hard to see exactly when the current becomes zero, so the stopping voltage can be a little wrong.
Assess your own answer
Tick the points your own answer already contains.
Read Planning a measurement in Investigate and compare.
Predict
Make your predictions before investigating 4 questions
Prediction 1: a brighter LED at the stopping voltage
Prediction 2: a stronger magnetic field
Prediction 3: one photon at a time
Optional Prediction 4: a hotter filament
Investigate
Before the visit: the physics of each station
Fotoni is the open physics teaching laboratory of the University of Helsinki. Its material bank describes the stations below. Your teacher chooses with Fotoni which ones your group does, and the equipment on the day can differ a little from the instructions: always check with the instructor.

Photoelectric effect: Planck's constant
LEDs of different colours shine on the metal plate of a vacuum tube (a photocell). A counter-voltage pushes the freed electrons back. You raise it until the current is zero: this is the stopping voltage $U_\text{s}$. From the unit Light as particles: the photoelectric effect you know that ${eU_\text{s} = hf - W}$. You measure $U_\text{s}$ for every LED and find Planck's constant $h$ from the slope of a graph.
The charge-to-mass ratio of the electron
A heated cathode emits electrons, and an accelerating voltage $U$ of about $100\text{–}300\ \text{V}$ speeds them up. The tube contains a little helium, so the beam glows. Two coils make a nearly uniform magnetic field $\overline{B}$, and the beam bends into a circle of radius $r$, as in the unit Charged particles and magnetic confinement.
The magnetic force is the centripetal force, and the voltage gives the kinetic energy:
$$\begin{aligned} evB &= \frac{mv^2}{r} \\ eU &= \tfrac{1}{2}mv^2 \end{aligned}$$
Together these give ${2U = (e/m) \cdot B^2r^2}$. The field of the coil pair is ${B = (4/5)^{3/2}\,\mu_0 N I / R}$, where $N$ is the number of turns of one coil, $R$ its radius and $I$ the current.
Double slit, also with single photons
Light passes two narrow slits, and a detector moves across the pattern. Bright light is measured with a photodiode. Then a very dim green lamp is used, so that there is practically never more than one photon in the apparatus. A photomultiplier turns each photon into a pulse, and a counter counts the pulses in $10\ \text{s}$. This links to the unit Waves, probability and quantum tunnelling.
Hot filament and spectra
The light of a lamp filament is spread out by a prism, and a sensor measures the intensity at each wavelength at four voltages ($7$, $8$, $9$ and $10\ \text{V}$). The software calculates the filament's temperature. Compare with Wien's law ${\lambda_\text{max} T = b}$ (unit When classical physics ran out of answers). At the spectra station you look at lamps and gas tubes with a hand spectroscope and a spectrometer (unit From Rutherford's nucleus to Bohr's atom).
Millikan's oil drops (qualitative)
Tiny oil drops float between two plates with a voltage of about $500\ \text{V}$. An alpha source charges them. You watch the drops through a microscope camera and look for sudden changes in their motion.
Which station measures what?
Check that you know what each station is for.
Planning a measurement
Every station has a research question: what do you want to find out? Then name the variables:
- the independent variable is the one you change (the LED, the coil current);
- the dependent variable is the one you measure (the stopping voltage, the radius);
- the controlled variables are kept the same (the accelerating voltage, the intensity).
No measurement is exact. The uncertainty $\Delta x$ of a quantity $x$ tells how far the true value may be from your reading. For a scale or a digital display, the reading uncertainty is about half of the smallest step you can read, unless the reading jumps around more. If you repeat a measurement, use the mean. Half of the range, (largest minus smallest) divided by two, is a simple estimate of how much single readings spread; use it as $\Delta x$ of one reading. It is not the uncertainty of the mean, which is smaller. Write the result as ${x \pm \Delta x}$ with its unit. The relative uncertainty $\Delta x / x$ is often given in percent.
Safety at the stations
- Several stations use high voltages: the electron tube up to $300\ \text{V}$, the oil-drop plates about $500\ \text{V}$, and the gas tubes. Change the connections only when the instructor allows it.
- The electron tube is thin glass: a magnet may come near it, but never touch it.
- Lamps and gas tubes get hot. Do not touch them, and do not look straight into a bright lamp or LED from close by.
- The alpha source of the oil-drop station stays inside the apparatus. Only the instructor handles it.
- Do not run the filament lamp at $10\ \text{V}$ for long: it shortens the lamp's life.
Safety check
Word bank
research question · independent variable · dependent variable · controlled variable · reading · uncertainty $\Delta x$ · to repeat · to keep constant · smallest step of the scale
Sentence starters
- Our research question is …
- We will change … and measure …
- We will keep … the same, because …
- The uncertainty of one reading is about …, because …
Plan the photoelectric measurement
Discuss with a partner and write a short plan for the photoelectric station: the research question, the independent, dependent and controlled variables, how many LEDs you will measure and how you will estimate the uncertainty of one stopping voltage.
Pair work: plan together, then each of you writes the plan on your own worksheet.
Model answer
Our research question is how the stopping voltage depends on the frequency of the light, and what Planck's constant is. We will change the LED, so the wavelength and the frequency are the independent variable. We will measure the stopping voltage, the dependent variable. We will keep the photocell and the metal plate the same. We will measure every LED the station has, at least four, because a line needs several points. The uncertainty of one stopping voltage is about one or two steps of the voltmeter's last digit, because it is hard to see exactly when the current reaches zero; we will find each stopping voltage twice and compare.
Assess your own answer
Tick the points your own answer already contains.
Read Planning a measurement again in Investigate.
During the visit: lab worksheet
Print this section, or copy the tables into your notebook. Fill in the empty cells at the station. Write every number with its unit and its uncertainty.
Group
| Names | Date | Stations we did |
|---|---|---|
Station: photoelectric effect
Research question: how does the stopping voltage depend on the frequency of the light, and what is Planck's constant?
| Instrument at the station | What it measures | Smallest step | $\Delta x$ |
|---|---|---|---|
| voltage $U$ | |||
| current $I$ |
Observations with the red LED, before you take the measurements:
| What we did | What happened to the current |
|---|---|
| $U$ slightly negative, more light | |
| $U$ at the stopping voltage, more light |
| LED colour | $\lambda\ (\text{nm})$ | $f\ (10^{14}\ \text{Hz})$ | $U_\text{s}\ (\text{V})$ |
|---|---|---|---|
| red | |||
Where does the LED's wavelength come from (a label, the instructor, a spectrometer)? Write it down.
Station: charge-to-mass ratio of the electron
Research question: what is the ratio $e/m$ of the electron?
| Coil data | Value | Where it comes from |
|---|---|---|
| turns of one coil $N$ | ||
| coil radius $R\ (\text{m})$ |
Observations before the measurements:
| What we did | What happened to the beam |
|---|---|
| a bar magnet near the tube (not touching) | |
| $U$ kept the same, coil current up | |
| coil current kept the same, $U$ up |
Measurements:
| $I\ (\text{A})$ | $U\ (\text{V})$ | $r\ (\text{m})$ |
|---|---|---|
Calculated from each row of the table above:
| $B\ (\text{T})$ | $B^2r^2\ (\text{T}^2\text{m}^2)$ | $2U\ (\text{V})$ |
|---|---|---|
How did you measure the radius, and what is $\Delta r$?
Station: double slit and single photons
Research question: does a pattern of bright and dark bands appear when photons arrive one at a time?
| Data of the setup | Value | Where it comes from |
|---|---|---|
| slit separation $d\ (\text{mm})$ | ||
| slit to detector $L\ (\text{cm})$ | ||
| wavelength $\lambda\ (\text{nm})$ |
Repeated counts at one position (each in $10\ \text{s}$):
| Count 1 | Count 2 | Count 3 | Count 4 | Count 5 |
|---|---|---|---|---|
Counts at different positions:
| Position $x\ (\text{mm})$ | Counts in $10\ \text{s}$ | Position $x\ (\text{mm})$ | Counts in $10\ \text{s}$ |
|---|---|---|---|
Station: hot filament
Research question: how does the spectrum of a hot filament change with its temperature?
| Voltage $U\ (\text{V})$ | Temperature $T\ (\text{K})$ | Peak $\lambda_\text{max}\ (\text{nm})$ | Colour we saw |
|---|---|---|---|
| $7$ | |||
| $8$ | |||
| $9$ | |||
| $10$ |
Station: spectra
Research question: which light sources give a continuous spectrum, and which give lines?
| Light source | Continuous or lines? | Brightest colours or lines |
|---|---|---|
Station: oil drops
Research question: how do charged oil drops move between the plates, and what happens when a drop's charge changes?
| What we saw | Our explanation |
|---|---|
After each station
One sentence: what surprised us, and what limited the precision?
| Station | Our sentence |
|---|---|
Language for your observations
Word bank
to increase · to decrease · to stay the same · to bend · to glow · a reading · a count · a peak · a line (in a spectrum) · continuous · roughly · exactly
Sentence starters
- When we increased …, the … stayed / grew / fell.
- The reading jumped between … and …, so …
- We repeated the count five times and got …
- We could not measure … because …
Theory summary
Lesson 2: after visit After the visit: open in lesson 2
After the visit: what the stations show
Compare these results with your own worksheet.
Photoelectric effect. When the counter-voltage is slightly negative, more intensity gives more current, because more photons free more electrons. At the stopping voltage the current stays at zero however bright the LED is. One photon frees at most one electron, and the photon's energy $hf$ depends only on the colour. Shorter wavelengths give larger stopping voltages, and the points $(f, U_\text{s})$ lie close to a straight line with the slope $h/e$.
Electron beam. A larger coil current makes a stronger field and a smaller circle, because ${r = mv/(eB)}$. A larger accelerating voltage makes the electrons faster and the circle larger. The points $(B^2r^2, 2U)$ lie on a line through the origin, and its slope is $e/m$, about ${1.76 \cdot 10^{11}\ \text{C/kg}}$.
Single photons. The counter clicks at random moments, and repeated counts at one place differ a little. But the counts at different positions form the same bright and dark bands as bright light does. Each photon arrives at one point like a particle, but the probability of where it arrives follows a wave pattern.
Hot filament. When the temperature rises, the peak moves to shorter wavelengths and the light looks whiter. The peak still stays in the infrared, so most of the energy of an ordinary filament lamp is not visible light. The curves fall at short wavelengths, as Planck's law says, not as classical physics predicted.
Spectra and oil drops. A filament gives a continuous spectrum. A gas tube gives lines at fixed wavelengths, for hydrogen $656$, $486$, $434$ and $410\ \text{nm}$. An oil drop that suddenly changes its speed has gained or lost charge. Millikan's measurements showed that the charge changes in whole steps of $e$.
Key idea
Each station shows a result that classical physics could not explain. Light gives and takes energy in quanta $hf$: the photoelectric effect, line spectra and the shape of the filament's spectrum all need Planck's constant. Electrons are particles with a fixed charge $e$ and a fixed ratio $e/m$. And single photons behave like particles when they are detected and like waves when they travel. This is the wave–particle duality of modern physics.
The same physics is a tool in fusion research. A charged particle circles around a magnetic field line and slides along it. Your electron beam shows the circling part, because it moves perpendicular to the field. A tokamak bends the field lines into closed loops inside the vessel, so the hot plasma stays away from the walls. Spectrometers measure the light of the plasma: at ITER, spectroscopic instruments cover the range from visible light to X-rays and measure, for example, which impurities the plasma contains and the temperature of its ions.
Back to your predictions
Compare your predictions with your own worksheet and with the results above. Which prediction did the measurement change most? Then answer the opening question in one or two sentences: what can we measure about light and electrons ourselves?
Would you change your prediction?
Words of the lab
Apply
A straight-line fit
When a law has the form $y = kx + b$, plot the measured points $(x, y)$ and fit the straight line that passes closest to all of them. A spreadsheet or a graphing calculator does this (a trendline or linear fit). For the photocell, ${U_\text{s} = (h/e)\,f - W/e}$, so
$$\begin{aligned} k &= \frac{h}{e}, & h &= e\,k, \\ b &= -\frac{W}{e}, & W &= -e\,b. \end{aligned}$$
For the electron beam, ${2U = (e/m) \cdot B^2r^2}$ is a line through the origin, and its slope is $e/m$.
The uncertainty of the slope can be estimated with the steepest and least steep lines that still pass through the uncertainty ranges of the points, usually the first and the last point. Then
$$\Delta k \approx \frac{k_\text{max} - k_\text{min}}{2}$$
Finally, compare your result with the accepted value by the percentage difference:
$$\text{difference} = \dfrac{x - x_\text{acc}}{x_\text{acc}} \cdot 100\ \%$$
where $x$ is your result and $x_\text{acc}$ the accepted value.
If the accepted value lies inside ${x \pm \Delta x}$, your result agrees with it within the uncertainty.
Worked example
Problem
A group measured the stopping voltage with five LEDs. The numbers are an invented example, rounded like real readings. The group estimated the uncertainty of every stopping voltage as $\Delta U_\text{s} = 0.02\ \text{V}$. Find Planck's constant with its uncertainty, and compare it with the accepted value ${6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$.
| LED | $\lambda\ (\text{nm})$ | $f\ (10^{14}\ \text{Hz})$ | $U_\text{s}\ (\text{V})$ |
|---|---|---|---|
| red | 625 | 4.80 | 0.55 |
| yellow | 590 | 5.08 | 0.69 |
| green | 525 | 5.71 | 0.93 |
| blue | 470 | 6.38 | 1.17 |
| violet | 405 | 7.40 | 1.60 |
The graph shows the five points and the line that a spreadsheet fits through them.
Step 1/7
Given: five pairs $(f, U_\text{s})$ from the table, ${\Delta U_\text{s} = 0.02\ \text{V}}$, the elementary charge ${e = 1.602 \cdot 10^{-19}\ \text{C}}$. We want Planck's constant $h$, its uncertainty $\Delta h$ and the percentage difference from ${6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$.
Step 2/7
Principle: Einstein's equation ${eU_\text{s} = hf - W}$ is a straight line ${U_\text{s} = (h/e)\,f - W/e}$. A fit through all five points gives the slope $k = h/e$ and uses every measurement. The steepest and least steep lines through the first and the last point, each moved by $\Delta U_\text{s}$, give the uncertainty of the slope.
Step 3/7
Formula: Solved in symbols, with the first point $(f_1, U_1)$ and the last point $(f_5, U_5)$:
$$\begin{aligned} h &= e\,k \\ k_\text{max} &= \frac{(U_5 + \Delta U_\text{s}) - (U_1 - \Delta U_\text{s})}{f_5 - f_1} \\ k_\text{min} &= \frac{(U_5 - \Delta U_\text{s}) - (U_1 + \Delta U_\text{s})}{f_5 - f_1} \\ \Delta h &= e \cdot \frac{k_\text{max} - k_\text{min}}{2} \end{aligned}$$
Step 4/7
Unit conversions: The frequencies are in units of $10^{14}\ \text{Hz}$, so the difference of the last and the first frequency is ${f_5 - f_1 = (7.40 - 4.80) \cdot 10^{14}\ \text{Hz}}$, that is, ${2.60 \cdot 10^{14}\ \text{Hz}}$. A slope in ${\text{V/Hz} = \text{V} \cdot \text{s}}$ times a charge in coulombs gives ${\text{C} \cdot \text{V} \cdot \text{s} = \text{J} \cdot \text{s}}$.
Step 5/7
Substitution: A spreadsheet fit of the five points gives the slope ${k = 3.964 \cdot 10^{-15}\ \text{V} \cdot \text{s}}$ and the intercept ${b = -1.34\ \text{V}}$. Then:
$$\begin{aligned} h &= 1.602 \cdot 10^{-19}\ \text{C} \\ &\quad \cdot 3.964 \cdot 10^{-15}\ \text{V} \cdot \text{s} \\ &= 6.351 \cdot 10^{-34}\ \text{J} \cdot \text{s} \\[6pt] k_\text{max} &= \frac{1.62\ \text{V} - 0.53\ \text{V}}{2.60 \cdot 10^{14}\ \text{Hz}} \\ &= 4.192 \cdot 10^{-15}\ \text{V} \cdot \text{s} \\[6pt] k_\text{min} &= \frac{1.58\ \text{V} - 0.57\ \text{V}}{2.60 \cdot 10^{14}\ \text{Hz}} \\ &= 3.885 \cdot 10^{-15}\ \text{V} \cdot \text{s} \\[6pt] \Delta h &= 1.602 \cdot 10^{-19}\ \text{C} \\ &\quad \cdot \frac{0.307 \cdot 10^{-15}\ \text{V} \cdot \text{s}}{2} \\ &= 0.246 \cdot 10^{-34}\ \text{J} \cdot \text{s} \\[6pt] \text{difference} &= \frac{6.351 - 6.626}{6.626} \\ &\quad \cdot 100\ \% \\ &= -4.2\ \% \end{aligned}$$
Step 6/7
Answer: ${h = (6.4 \pm 0.3) \cdot 10^{-34}\ \text{J} \cdot \text{s}}$. The uncertainty is rounded up to one significant figure, and the value is given to the same decimal place. The result is about $4\ \%$ smaller than the accepted value.
Step 7/7
Reasonableness: The accepted value ${6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$ lies inside ${(6.1\ \text{to}\ 6.7) \cdot 10^{-34}\ \text{J} \cdot \text{s}}$, so the result agrees with it within the uncertainty. The intercept gives ${W = -e\,b \approx 1.34\ \text{eV}}$, positive and of the order of one electronvolt, as a work function should be. Possible systematic errors: an LED gives a band of wavelengths, not one exact wavelength, and it is hard to decide when the current is exactly zero.
Optional The charge-to-mass ratio from one reading
Example data: at $U = 200\ \text{V}$ and a coil current of $I = 1.50\ \text{A}$, a group measured the diameter of the circle as $8.2\ \text{cm}$. One coil has $N = 130$ turns and the radius $R = 0.150\ \text{m}$, as in the Fotoni instruction. Calculate $e/m$. Give the answer as a number times $10^{11}\ \text{C/kg}$: for example, type $1.9$ for ${1.9 \cdot 10^{11}\ \text{C/kg}}$.
Hint 1: a guiding question
First find the field: $B = (4/5)^{3/2}\,\mu_0 N I / R$, with ${\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}}$.
Hint 2: what is the connection
The equation is ${2U = (e/m) \cdot B^2r^2}$, so $e/m = 2U/(B^2r^2)$. Is $8.2\ \text{cm}$ the radius?
Hint 3: an intermediate step
${B \approx 1.169 \cdot 10^{-3}\ \text{T}}$ and $r = 0.041\ \text{m}$, so $B^2r^2 \approx 2.30 \cdot 10^{-9}\ \text{T}^2 \cdot \text{m}^2$.
Show the solution
Given
accelerating voltage $U = 200\ \text{V}$, coil current $I = 1.50\ \text{A}$, diameter of the circle $8.2\ \text{cm}$, turns of one coil $N = 130$, coil radius $R = 0.150\ \text{m}$, ${\mu_0 = 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A}}$. We want $e/m$.
Principle
The magnetic force is the centripetal force, $evB = mv^2/r$, and the accelerating voltage gives the kinetic energy, $eU = \tfrac{1}{2}mv^2$. Eliminating $v$ gives ${2U = (e/m) \cdot B^2r^2}$. The field of the coil pair is $B = (4/5)^{3/2}\,\mu_0 N I / R$.
Formula
Solved in symbols:
$$\begin{aligned} B &= \left(\frac{4}{5}\right)^{3/2} \frac{\mu_0 N I}{R} \\ \frac{e}{m} &= \frac{2U}{B^2 r^2} \end{aligned}$$
Unit conversions
The radius is half of the diameter, in metres:
The units: ${\text{V}/(\text{T}^2 \cdot \text{m}^2) = \text{C/kg}}$.
Substitution
$$\begin{aligned} \mu_0 N I &= 4\pi \cdot 10^{-7}\ \text{T} \cdot \text{m/A} \\ &\quad \cdot 130 \cdot 1.50\ \text{A} \\ &= 2.450 \cdot 10^{-4}\ \text{T} \cdot \text{m} \\[6pt] B &= 0.7155 \cdot \frac{2.450 \cdot 10^{-4}\ \text{T} \cdot \text{m}}{0.150\ \text{m}} \\ &= 1.169 \cdot 10^{-3}\ \text{T} \\[6pt] B^2r^2 &= (1.169 \cdot 10^{-3}\ \text{T})^2 \\ &\quad \cdot (0.041\ \text{m})^2 \\ &= 2.297 \cdot 10^{-9}\ \text{T}^2 \cdot \text{m}^2 \\[6pt] \frac{e}{m} &= \frac{2 \cdot 200\ \text{V}}{B^2r^2} \\ &= \frac{400\ \text{V}}{2.297 \cdot 10^{-9}\ \text{T}^2 \cdot \text{m}^2} \\ &= 1.741 \cdot 10^{11}\ \text{C/kg} \end{aligned}$$
Answer
${e/m \approx 1.74 \cdot 10^{11}\ \text{C/kg}}$
The diameter has only two significant figures, so a report would give ${1.7 \cdot 10^{11}\ \text{C/kg}}$; the third figure is kept here only to compare with the accepted value.
Reasonableness
The accepted value is ${1.76 \cdot 10^{11}\ \text{C/kg}}$, so the result is about $1\ \%$ too small. A radius error of only $1\ \text{mm}$ changes $r^2$ by about $5\ \%$, so the radius is the most important uncertainty, and one point is not enough: at Fotoni you measure six and fit a line.
Sentence starters
- We plotted … on the vertical axis and … on the horizontal axis.
- The fit gave a slope of …, so $h$ = …
- The steepest and least steep lines gave …, so $\Delta h$ ≈ …
- Our result is …% smaller / larger than the accepted value, which is inside / outside our uncertainty.
Analyse your own photoelectric data
Use the table on your worksheet (or the example table above, if your group did not do this station). Plot $U_\text{s}$ against $f$ in a spreadsheet, fit a line and find $h$, $\Delta h$ and the percentage difference from ${6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$. Write your result and three or four sentences on how you got it.
Model answer
We plotted the stopping voltage on the vertical axis and the frequency on the horizontal axis. The five points lay close to a straight line, and the spreadsheet fit gave the slope ${3.96 \cdot 10^{-15}\ \text{V} \cdot \text{s}}$, so ${h = e\,k = 6.35 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$. The steepest and least steep lines gave ${\Delta h \approx 0.3 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$, so ${h = (6.4 \pm 0.3) \cdot 10^{-34}\ \text{J} \cdot \text{s}}$. This is about $4\ \%$ smaller than the accepted value, which lies inside our uncertainty. (These are the numbers of the example table; your own will differ.)
Assess your own answer
Tick the points your own answer already contains.
Follow the worked example step by step in Apply.
Hint 1: a guiding question
The frequency is $f = c/\lambda$ with ${c = 2.998 \cdot 10^{8}\ \text{m/s}}$ and $\lambda$ in metres.
Hint 2: what is the connection
In most spreadsheets: make a scatter chart, add a linear trendline and show its equation. The slope is in ${\text{V/Hz} = \text{V} \cdot \text{s}}$.
Hint 3: an intermediate step
For $\Delta h$, draw the steepest and least steep lines through the first and last point moved by ${\pm \Delta U_\text{s}}$, as in the worked example.
Optional The spacing of the bands
In the Fotoni double-slit setup the slit separation is $d = 0.3556\ \text{mm}$ and the detector is $L = 49\ \text{cm}$ from the slits. For small angles, neighbouring bright bands are $\Delta y = \lambda L / d$ apart.
How far apart are they for green light of $550\ \text{nm}$? Give the answer in millimetres.
Hint 1: a guiding question
Write all three lengths in metres before you substitute.
Hint 2: what is the connection
${\lambda = 550 \cdot 10^{-9}\ \text{m}}$, $L = 0.49\ \text{m}$ and ${d = 0.3556 \cdot 10^{-3}\ \text{m}}$.
Show the solution
Given
$\lambda = 550\ \text{nm}$, $L = 49\ \text{cm}$, $d = 0.3556\ \text{mm}$. We want the distance $\Delta y$ between neighbouring bright bands.
Principle
Bright bands appear where the paths from the two slits differ by a whole number of wavelengths, $d \sin\theta_n = n\lambda$. For small angles $\sin\theta \approx \tan\theta = y/L$, so the bands are evenly spaced.
Formula
$y_n = n\lambda L/d$, so neighbouring bands are $\Delta y = \lambda L/d$ apart.
Unit conversions
${\lambda = 550 \cdot 10^{-9}\ \text{m}}$, $L = 0.49\ \text{m}$, ${d = 0.3556 \cdot 10^{-3}\ \text{m}}$.
Substitution
$$\begin{aligned} \Delta y &= \frac{550 \cdot 10^{-9}\ \text{m} \cdot 0.49\ \text{m}}{0.3556 \cdot 10^{-3}\ \text{m}} \\ &= 7.58 \cdot 10^{-4}\ \text{m} \\ &= 0.758\ \text{mm} \end{aligned}$$
Answer
$\Delta y \approx 0.76\ \text{mm}$
Two significant figures, as in $L = 49\ \text{cm}$.
Reasonableness
The bands are less than a millimetre apart, so the detector needs a narrow measuring slit that moves in small steps. The angle is about $0.76\ \text{mm} : 490\ \text{mm} \approx 0.0015$, so the small-angle approximation is very good.
Word bank
quantum (plural quanta) · photon · Planck's constant · wave–particle duality · probability · charge-to-mass ratio · magnetic confinement · spectroscopy · impurity · classical physics
Sentence starters
- Classical physics predicted …, but we saw …
- This result is evidence that …
- The same idea is used in fusion research, where …
- The experiment was important because …
Connect to modern physics
Discuss in a group of three. Choose two stations you did. For each one, write two sentences: what the measurement shows that classical physics could not explain, and how the same physics is used in fusion research or elsewhere in modern technology.
Group work in threes: each of you explains one station to the others, then each writes an own answer.
Model answer
At the photoelectric station the stopping voltage did not change when the LED became brighter, only when its colour changed. Classical physics predicted that brighter light gives faster electrons, so this is evidence that light comes in quanta of energy $hf$. At the electron-beam station the electrons moved on a circle whose radius depended on the field, $r = mv/(eB)$. The same motion of charged particles around magnetic field lines is the idea of a tokamak, which keeps a hot plasma away from its walls.
Assess your own answer
Tick the points your own answer already contains.
Read the key idea in the Theory summary.
Writing a lab report in English
A lab report tells a reader what you did, what you found and how sure you are. Scientists write it in a fixed order:
- Title and aim: the research question in one sentence.
- Background: the physics and the equation you test, in the present tense ("The stopping voltage depends on …").
- Method: the equipment and what you did, in the past tense ("We measured …", or the passive "The voltage was increased …").
- Results: tables with units in the headings, a graph with labelled axes and the fitted line.
- Analysis: the fit, the result with its uncertainty, ${x \pm \Delta x}$.
- Discussion: the comparison with the accepted value, the sources of uncertainty and how to improve the experiment.
- Conclusion: the answer to the research question in one or two sentences.
Write numbers with a decimal point and their unit ($0.93\ \text{V}$). Do not claim more than the data shows: "the result agrees with the accepted value within its uncertainty" is stronger science than "the result proves the theory".
Put the report in order
Word bank
accepted value · percentage difference · within the uncertainty · source of uncertainty · systematic error · random error · to improve · to repeat · to agree with · to be consistent with
Sentence starters
- The evidence suggests that …
- Our value agrees with the accepted value within …
- The main source of uncertainty was …, because …
- A limitation of our measurement is …
- To improve the experiment, we could …
Write the discussion of your report
Write the Discussion of your photoelectric report (or of another quantitative station) in five to seven sentences. Compare your result with the accepted value, name the main sources of uncertainty, say whether they are random or systematic, and suggest one improvement.
Model answer
Our value ${h = (6.4 \pm 0.3) \cdot 10^{-34}\ \text{J} \cdot \text{s}}$ is about $4\ \%$ smaller than the accepted value ${6.626 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$, but the accepted value lies within our uncertainty. The main source of uncertainty was deciding when the current was exactly zero, because the reading jumped between two values; this is a random error. A possible systematic error is that an LED does not give one exact wavelength but a band of wavelengths, so the frequencies in our table are not exact. A limitation of our measurement is that we had only five LEDs. To improve the experiment, we could measure each stopping voltage several times and find the wavelength of each LED with a spectrometer.
Assess your own answer
Tick the points your own answer already contains.
Read Writing a lab report in English in Apply and use the word bank.
Common misconception
"An uncertainty means that we made a mistake." No: every measurement has an uncertainty, even a perfect one. A mistake (a wrong connection, a misread scale) is something you can find and correct. The uncertainty estimates how far the true value may be from the result. Part of it shows as scatter when you repeat the measurement. Systematic errors, such as the band of wavelengths of an LED, do not show as scatter, and repeating does not reduce them.
A second misconception is that more decimals mean a better result. A calculator may show ${6.351596 \cdot 10^{-34}}$, but with ${\Delta h = 0.3 \cdot 10^{-34}\ \text{J} \cdot \text{s}}$ only the first decimal means something. A third is that a result that differs from the accepted value is wrong. It is fine if the accepted value lies inside ${x \pm \Delta x}$.
Check your understanding
1. Concept: does the result agree?
2. Graph: which axes for e/m?
3. Data: report a repeated count
Sentence starters
- I disagree, because a difference of … is …
- An experiment fails only if …
- Every measurement has …
- To make the result more reliable, we could …
4. Explain and argue
A classmate writes in a report: "Our value of $e/m$ was ${1.74 \cdot 10^{11}\ \text{C/kg}}$, $1\ \%$ smaller than the accepted value, so the experiment failed." Do you agree? Explain the difference between an uncertainty and a mistake, and suggest two ways to make the result more reliable.
Model answer
I disagree. Every measurement has an uncertainty, and a difference of $1\ \%$ is small: a radius error of only $1\ \text{mm}$ in $4\ \text{cm}$ already changes $e/m$ by about $5\ \%$. An experiment fails only if the accepted value lies far outside ${x \pm \Delta x}$, or if there was a mistake such as a wrong connection or using the diameter instead of the radius. To make the result more reliable, we could measure the radius several times and use the mean, and measure at several voltages and currents and find $e/m$ from the slope of a fitted line.
Assess your own answer
Tick the points your own answer already contains.
Read Common misconception in Apply.
Summary
- At Fotoni you measure modern physics yourself: Planck's constant from stopping voltages, $e/m$ from a circular electron beam, single photons in a double slit, the spectrum of a hot filament, line spectra and charged oil drops.
- Plan every measurement: research question, independent, dependent and controlled variables, and the uncertainty $\Delta x$ of each reading.
- The slope of a fitted line gives the constant: ${h = e\,k}$ from ${U_\text{s} = (h/e)\,f - W/e}$, and $e/m$ from ${2U = (e/m) \cdot B^2r^2}$.
- The steepest and least steep lines give ${\Delta k \approx (k_\text{max} - k_\text{min})/2}$. Report ${x \pm \Delta x}$ and the percentage difference from the accepted value.
- A lab report has a fixed structure: aim, background, method, results, analysis, discussion and conclusion.
Le coin français
Many words of the lab are almost the same in French: le laboratoire, une mesure, un spectre. A French lab report also discusses l'incertitude. Listen, repeat, and then tell your partner in French what you measured at Fotoni. In the hints, capitals show the stressed syllable, and õ, ã and ẽ are nasal vowels: the air goes through the nose and no n is heard.
Photographs and videos
- Fotoni logo. University of Helsinki, Fotoni. source, © University of Helsinki (logo; not freely licensed). Use only to identify the Fotoni lab, unmodified.