F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsInvestigation: Measuring Planck's Constant
F2.4 F2.2 F3.1About 70 minutes
Learning goals- I can describe how a photoelectric experiment and an LED experiment can be used to measure Planck's constant.
- I can plot data, draw a line of best fit and use the slope and intercepts to find h, the work function and the threshold frequency.
- I can calculate percent error and identify sources of error in quantum experiments.
- I can explain how the data support the photon model of light.
Words to know
line of best fit
A straight line drawn (or calculated) so that the data points are spread as evenly as possible on both sides of it.
percent error
|experimental − accepted| ÷ accepted × 100%.
random error
Unpredictable scatter in measurements that makes repeated readings differ; reduced by repeating and averaging.
systematic error
A consistent bias that shifts every measurement in the same direction; not removed by averaging.
light-emitting diode (LED)
A semiconductor device that gives off photons when electrons drop across an energy gap.
energy gap
In a semiconductor, the energy difference that electrons drop through when they emit a photon in an LED.
Let the data decide
A model in physics is only as good as the evidence for it. Einstein's equation makes a sharp, testable prediction: plot E(k,max) (or the stopping voltage) against frequency and you must get a straight line whose slope is Planck's constant, the same number Planck found from glowing objects. In this lesson you will analyse data the way Millikan did, and look at a second, simpler classroom method that uses LEDs.
This lesson practises the scientific investigation skills you use throughout the course: planning, recording, graphing, analysing and evaluating.
Method 1: the photoelectric (h/e) apparatus
Many school labs have a phototube and a mercury lamp. A mercury lamp gives off light at a few sharp wavelengths, including 577 nm (yellow), 546 nm (green), 436 nm (blue), 405 nm (violet) and 365 nm (ultraviolet). Filters or a small grating select one line at a time. For each line, the reverse voltage is increased until the photocurrent just drops to zero; that voltage is V₀.
Safety: mercury lamps emit ultraviolet light that can harm the eyes and skin. Never look into the lamp; keep it in its housing; let it cool before handling. This is a school-lab or simulation investigation, not a home one.
Since eV₀ = hf − W, dividing by e gives
V₀ = (h/e) f − W/e
A graph of V₀ against f has slope h/e, horizontal intercept f₀ and vertical intercept −W/e.
Worked exampleExample 1: Analysing photoelectric data
A class records these results.
- λ (nm): 577, 546, 436, 405, 365
- V₀ (V): 0.26, 0.36, 0.95, 1.15, 1.51
Find h, the work function and the threshold frequency, and the percent error in h (accepted value 6.63 × 10⁻³⁴ J·s).
- Convert to frequency with f = c/λ: 5.20, 5.49, 6.88, 7.41 and 8.22 (all × 10¹⁴ Hz).
- Plot V₀ against f and draw the line of best fit (graph below). Two points far apart on the line (not data points) are (5.20 × 10¹⁴ Hz, 0.25 V) and (8.22 × 10¹⁴ Hz, 1.50 V).
- Slope = (1.50 − 0.25) V ÷ (3.02 × 10¹⁴ Hz) = 4.14 × 10⁻¹⁵ V/Hz. (A calculator's least-squares fit gives 4.15 × 10⁻¹⁵ V/Hz.)
- h = slope × e = (4.15 × 10⁻¹⁵)(1.60 × 10⁻¹⁹) = 6.64 × 10⁻³⁴ J·s.
- Vertical intercept: the line meets f = 0 at about −1.91 V, so W = 1.91 eV. Horizontal intercept: f₀ = W/h = 1.91 eV ÷ 4.15 × 10⁻¹⁵ eV·s = 4.6 × 10¹⁴ Hz.
- Percent error = |6.64 − 6.63| ÷ 6.63 × 100% = 0.2%.
h ≈ 6.6 × 10⁻³⁴ J·s (0.2% error), W ≈ 1.9 eV, f₀ ≈ 4.6 × 10¹⁴ Hz (λ₀ ≈ 650 nm). The five points lie very close to a straight line, just as Einstein's equation predicts. (Real student results are usually less exact than this example; within 10% is good.)
Stopping voltage against frequency for the five mercury lines in Example 1. The points fall on a straight line; its slope times e gives Planck's constant, and it crosses the axis at the threshold frequency.Method 2: LEDs
A light-emitting diode is a semiconductor. When a voltage pushes current through it, electrons drop across an energy gap and each drop releases one photon with energy about equal to the gap. An LED barely conducts until the voltage across it supplies roughly that much energy per electron. So for each LED colour,
eV(turn-on) ≈ hf = hc/λ, which gives V(turn-on) ≈ (hc/e)(1/λ)
Measure the turn-on voltage for several colours, plot V against 1/λ, and the slope is about hc/e. Real LEDs emit a small range of wavelengths and have no perfectly sharp turn-on, so this method usually gives h within about 10%.
In an LED, each electron that drops across the energy gap gives off one photon. The voltage needed to start the current is roughly the gap energy divided by e, so blue LEDs need more voltage than red ones.Worked exampleExample 2: Planck's constant from LEDs
Turn-on voltages for five LEDs:
- λ (nm): 940 (infrared), 630 (red), 590 (yellow), 525 (green), 470 (blue)
- V (V): 1.25, 1.88, 2.02, 2.30, 2.58
Find h.
- Calculate 1/λ: 1.06, 1.59, 1.69, 1.90 and 2.13 (all × 10⁶ m⁻¹).
- Plot V against 1/λ. The best-fit line passes close to (1.06 × 10⁶ m⁻¹, 1.25 V) and (2.13 × 10⁶ m⁻¹, 2.58 V).
- Slope = (2.58 − 1.25) ÷ (2.13 − 1.06) × 10⁶ = 1.33 ÷ 1.07 × 10⁶ = 1.24 × 10⁻⁶ V·m.
- Slope = hc/e, so h = slope × e ÷ c = (1.24 × 10⁻⁶)(1.60 × 10⁻¹⁹) ÷ (3.00 × 10⁸) = 6.6 × 10⁻³⁴ J·s.
- Percent error ≈ |6.6 − 6.63| ÷ 6.63 × 100% ≈ 0.5%, which is unusually good for this method.
h ≈ 6.6 × 10⁻³⁴ J·s. Notice the colours line up in order: the shorter the wavelength, the higher the turn-on voltage, as the photon model predicts.
Sources of error to discuss in a lab report
- Deciding when the current is "zero" (photocell) or when the LED "turns on": both are judgement calls; a sensitive meter helps.
- Stray light in the room adds photocurrent and shifts V₀ (systematic). Work in a darkened room.
- Contact voltages in the phototube shift every V₀ by the same amount: this moves the intercept but not the slope, which is why h is found from the slope.
- LEDs emit a band of wavelengths, not one; the listed λ is only the peak.
- Too few points or points bunched together make the slope uncertain. Use the widest range of frequencies you can.
Just for funMillikan spent years trying to prove Einstein's photon equation wrong, and ended up with some of the best evidence that it was right. Being wrong can be very productive when you measure carefully.
Hands-on activityCollect and analyse your own photoelectric data
You need: a computer or tablet with a browser, the free PhET "Photoelectric Effect" simulation, and a spreadsheet (or graph paper).
- Pick a metal other than sodium (for example zinc or calcium). Set the intensity to a middle value.
- Choose at least five wavelengths that produce electrons, spread as widely as possible (for zinc, all must be shorter than about 290 nm).
- For each, adjust the battery voltage until the current just reaches zero. Record λ and V₀ in a table.
- Calculate f = c/λ for each. Plot V₀ (y) against f (x). Add a linear trendline and display its equation.
- Calculate h = slope × e, the work function W = −(intercept) in eV, and the percent error in h.
- Write a three-sentence conclusion: does your graph support Einstein's model? What was your biggest source of error?
What to look for: a straight line with slope near 4.1 × 10⁻¹⁵ V/Hz (h ≈ 6.6 × 10⁻³⁴ J·s) and a work function close to the value the simulation uses for that metal. If one point is far off the line, repeat that measurement: it was probably a reading slip.
Your turn
1Why do we find h from the slope of the V₀ vs f graph rather than from a single measurement?
A single point also involves W, which we may not know exactly. The slope uses the change in V₀ for a change in f, so W (and any constant offset such as contact voltages) cancels out. Using many points also averages out random errors.
2A best-fit line on a V₀ vs f graph has slope 3.9 × 10⁻¹⁵ V/Hz. Find h and the percent error.
h = (3.9 × 10⁻¹⁵)(1.60 × 10⁻¹⁹) = 6.2 × 10⁻³⁴ J·s. Percent error = |6.2 − 6.63| ÷ 6.63 × 100% ≈ 6%.
3The line in question 2 crosses the V₀-axis at −2.2 V. What is the work function and the threshold frequency?
W = 2.2 eV. f₀ = W ÷ slope = 2.2 V ÷ 3.9 × 10⁻¹⁵ V/Hz = 5.6 × 10¹⁴ Hz.
4A green LED (525 nm) turns on at 2.30 V. Use only this one LED to estimate h.
h ≈ eVλ/c = (1.60 × 10⁻¹⁹)(2.30)(525 × 10⁻⁹) ÷ (3.00 × 10⁸) = 6.4 × 10⁻³⁴ J·s.
5Classify each error as random or systematic: (a) the room lights were on during the whole experiment; (b) the needle on the meter wobbled and readings varied each time.
(a) Systematic (stray light shifts every reading the same way). (b) Random (unpredictable scatter; reduce by repeating and averaging).
6Why does a graph of V₀ vs f support the photon model rather than the wave model?
The wave model predicts no dependence of electron energy on frequency (only on intensity). The data show V₀ rising linearly with f with a slope equal to h/e, exactly as E(k,max) = hf − W predicts.
More practice: warm-up
1What quantity is on each axis of a Millikan-style graph?
Stopping voltage V₀ (or E(k,max)) on the y-axis; frequency f on the x-axis.
2What does the slope of V₀ against f equal?
3Convert a wavelength of 436 nm to frequency.
f = 3.00 × 10⁸ ÷ 4.36 × 10⁻⁷ = 6.88 × 10¹⁴ Hz
4Write the formula for percent error.
|experimental − accepted| ÷ accepted × 100%
5Which LED needs a higher turn-on voltage, red or blue?
Blue (higher-energy photons)
More practice: on your own
1A student found h = 7.1 × 10⁻³⁴ J·s. Find the percent error and suggest one reason the slope might be too steep.
Percent error = 0.47 ÷ 6.63 × 100% ≈ 7%. Possible reason: the stopping voltage was over-estimated more at high frequencies (for example, judging "zero" current differently for dimmer UV lines), or stray light affected some points more than others.
2Two points on a best-fit line are (6.0 × 10¹⁴ Hz, 0.50 V) and (9.0 × 10¹⁴ Hz, 1.75 V). Find h, W and f₀.
Slope = 1.25 V ÷ 3.0 × 10¹⁴ Hz = 4.17 × 10⁻¹⁵ V/Hz; h = 6.7 × 10⁻³⁴ J·s. Intercept: 0.50 − (4.17 × 10⁻¹⁵)(6.0 × 10¹⁴) = 0.50 − 2.50 = −2.0 V, so W = 2.0 eV. f₀ = 2.0 ÷ 4.17 × 10⁻¹⁵ = 4.8 × 10¹⁴ Hz.
3An infrared LED (940 nm) turns on at 1.25 V. Predict the turn-on voltage of a violet LED at 405 nm, assuming V is proportional to 1/λ.
V ∝ 1/λ, so V = 1.25 × (940 ÷ 405) = 2.9 V.
4Why should you choose mercury lines spread from 577 nm to 365 nm rather than five lines between 436 nm and 405 nm?
A wide range of x-values makes the slope much less sensitive to small errors in each V₀. Bunched points could allow very different lines to fit almost equally well.
5In Example 1, a contact voltage of 0.3 V added to every reading. Which results change: h, W or f₀?
Adding 0.3 V to every V₀ moves the line up without changing its slope. h is unchanged; the intercept changes, so the calculated W and f₀ change.
Challenge
1Design a short procedure to test the claim "the stopping voltage does not depend on light intensity," using the phototube. Include your independent, dependent and controlled variables.
Independent: intensity (change lamp distance or use neutral-density filters; or the intensity slider in a simulation). Dependent: stopping voltage V₀. Controlled: the wavelength (one mercury line), the metal, the tube, the room darkness. Procedure: for one line, measure V₀ at 4 or 5 intensities, repeating each 3 times; plot V₀ against intensity. Expected: a horizontal line (V₀ constant) while the saturation current rises with intensity.
2Using the data in Example 1, a student calculated h from just the 577 nm and 365 nm points. Show the calculation and compare with the best-fit value.
Slope = (1.51 − 0.26) ÷ (8.22 − 5.20) × 10¹⁴ = 1.25 ÷ 3.02 × 10¹⁴ = 4.14 × 10⁻¹⁵ V/Hz. h = 4.14 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.62 × 10⁻³⁴ J·s, almost the same as the best-fit 6.64 × 10⁻³⁴ J·s. Here the end points happen to be good; in general a best-fit line using all points is more reliable.
3The LED method relies on eV ≈ hc/λ. Give one physical reason this is only approximate, and explain whether it would make the calculated h too big or too small if the turn-on voltage is systematically read a little low.
Reasons: an LED emits a band of wavelengths; the current rises gradually so "turn-on" is not sharp; some energy is lost as heat in the diode. If every V is read a little low by the same fraction, the slope (hc/e) is too small, so h is too small. (A constant offset would change the intercept, not the slope.)
Check your understandingQuiz
1. In a graph of stopping voltage against frequency, Planck's constant is found from
2. A student's measured h is 6.30 × 10⁻³⁴ J·s. The percent error is about
3. Which change would most improve the reliability of a measured slope?
4. Stray room light raises every photocurrent reading in a phototube experiment. This is an example of
5. Why does a blue LED need a larger turn-on voltage than a red LED?
6. The vertical intercept of a V₀ against f graph is −2.1 V. The work function is
For parents and teachersBig idea (F2.4): lab or simulation data can test a quantum theory. The straight-line graph of stopping voltage against frequency, with slope h/e, is strong evidence for Einstein's photon model.
How to help: a spreadsheet makes this lesson much smoother. Sit with your teen while they enter the data, add a trendline and interpret its slope and intercept. Ask them to explain the difference between a random and a systematic error with an everyday example (a bathroom scale that always reads 1 kg high is systematic).
Common difficulties:
- Using data points instead of points on the best-fit line to calculate the slope.
- Forgetting to convert wavelength to frequency (or to 1/λ for LEDs) before plotting.
- Forgetting the powers of ten when reading slopes off a graph whose axis says "× 10¹⁴ Hz."
- Writing "human error" as a source of error. Encourage specific causes: stray light, judging zero current, wavelength spread of an LED.
Safety: mercury lamps and UV sources belong in a supervised school lab. The home activity uses a simulation only.