Physics · Simulation and virtual lab Free
Photoelectric effect: stopping voltage and Planck’s constant
Shine light on a metal in a photocell, find the stopping voltage for each colour and measure Planck’s constant.
- You will learn to
- Show that electrons are emitted only above a threshold frequency.
- Measure stopping voltages and plot them against frequency.
- Determine Planck’s constant and the work function.
- You will measure
- Wavelength (nm), frequency (× 10¹⁴ Hz), stopping voltage (V), current (µA).
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: photocell, light source with filters, variable DC supply, voltmeter, microammeter.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate the photoelectric effect, to show that the maximum kinetic energy of the electrons depends on the frequency and not on the intensity of the light, and to determine Planck’s constant and the work function of a metal.
What to do
- Choose Caesium and a wavelength of 400 nm. Read the current with the anode voltage at 0 V.
- Make the voltage negative, slowly (0.01 V steps), until the current just becomes 0.00 µA. Tap Record the stopping voltage.
- Repeat for 450, 500, 550 and 580 nm. Look at the graph of stopping voltage against frequency.
- Find h = e × gradient and the threshold frequency from the intercept on the frequency axis.
- Drag the lamp nearer: what happens to the current? Does the stopping voltage change?
- Choose Zinc. Try visible light, then ultraviolet (below 290 nm). Explain the result.
Results
📏 Read the instruments carefully
| Instrument | Smallest reading | Uncertainty |
|---|
| Voltmeter | 0.01 V | ± 0.01 V |
| Microammeter | 0.01 µA | ± 0.01 µA |
| Filter wavelength | 5 nm | ± 5 nm (each filter passes a narrow band) |
Write every reading with its unit, and with the same number of decimal places as the instrument. Use the 🔍 Magnifier on the picture to read small scales at eye level.
⚠️ Avoid these mistakes
- Changing the voltage too quickly: the stopping voltage is where the current JUST becomes zero.
- Thinking brighter light gives faster electrons: intensity changes the current, not the stopping voltage.
- Forgetting to convert: 1 eV = 1.6 × 10⁻¹⁹ J, and f = c ÷ λ with λ in metres.
✍️ Conclusion
Write your own conclusion first: what do your results and graph show, and which law or principle do they support?
Check your conclusion
Electrons are emitted only when the photon energy hf is larger than the work function W. The maximum kinetic energy, eVs = hf − W, increases with frequency but not with intensity. The graph of Vs against f is a straight line of gradient h ÷ e, which gives Planck’s constant h ≈ 6.6 × 10⁻³⁴ J s; light behaves as particles (photons).
Questions
What is the photoelectric effect?
The emission of electrons from a metal surface when light (electromagnetic radiation) of high enough frequency falls on it.
What is the work function of a metal?
The minimum energy needed to remove an electron from the surface of the metal.
Why does red light not release electrons from zinc, however bright it is?
Each photon of red light has energy hf smaller than the work function of zinc; one photon gives its energy to one electron, so no electron can escape, whatever the number of photons.
What happens to the current when the lamp is brought closer? Why?
The current increases because more photons per second arrive, so more electrons are emitted per second; the stopping voltage does not change.
Write Einstein’s photoelectric equation.
hf = W + KEmax, where KEmax = eVs is the maximum kinetic energy of the emitted electrons.
Calculate the energy of a photon of wavelength 500 nm in eV.
E = hc ÷ λ = 6.63 × 10⁻³⁴ × 3.0 × 10⁸ ÷ 5.0 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J = 2.48 eV.
For teachers: expected results and how the model works
Photon energy = 1240 ÷ λ(nm) eV. Work functions: caesium 2.10 eV (threshold 590 nm), potassium 2.30 eV (539 nm), calcium 2.90 eV (428 nm), zinc 4.30 eV (288 nm). Saturation current = 1.5 µA × (20 cm ÷ d)². For a negative anode voltage the current falls linearly to zero at Vs. Learners obtain h ≈ 6.6 × 10⁻³⁴ J s.
Quick check 3 questions · scored
Choose one answer for each question. You see at once if it is right, why, and your score at the end.
1. Red light does not release electrons from zinc. What happens if the red light is made much brighter?
Each red photon still has too little energy; brightness only adds more photons of the same energy.
2. Photon energy 3.1 eV, work function 2.1 eV. Stopping voltage?
KEmax = 3.1 − 2.1 = 1.0 eV, so Vs = 1.0 V.
3. The Vs against f graph has a gradient of 4.1 × 10⁻¹⁵ V s. What is h?
h = e × gradient = 1.6 × 10⁻¹⁹ × 4.1 × 10⁻¹⁵ ≈ 6.6 × 10⁻³⁴ J s.
Your score:
Exam practice Exam set
National-exam style questions on this simulation (15 marks). Try each one on paper first, then open the marking scheme.
Light of frequency 7.5 × 10¹⁴ Hz falls on caesium (work function 2.1 eV). Calculate (a) the photon energy in eV, (b) the maximum kinetic energy of the electrons, (c) the stopping voltage. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C) (6 marks)
Marking scheme: (a) E = hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J (1) = 3.1 eV (1). (b) KEmax = 3.1 − 2.1 (1) = 1.0 eV (1). (c) Vs = KEmax ÷ e = 1.0 V (2).
A student plots stopping voltage against frequency and obtains a straight line of gradient 4.1 × 10⁻¹⁵ V s, cutting the frequency axis at 5.1 × 10¹⁴ Hz. Find Planck’s constant and the work function in eV. (4 marks)
Marking scheme: h = e × gradient (1) = 1.6 × 10⁻¹⁹ × 4.1 × 10⁻¹⁵ = 6.6 × 10⁻³⁴ J s (1). W = h f₀ (1) = 6.6 × 10⁻³⁴ × 5.1 × 10¹⁴ = 3.4 × 10⁻¹⁹ J = 2.1 eV (1).
State three observations of the photoelectric effect that the wave theory of light cannot explain. (3 marks)
Marking scheme: No emission below a threshold frequency, however intense the light (1); KEmax depends on frequency, not intensity (1); emission is instantaneous even with very dim light (1).
State two applications of the photoelectric effect. (2 marks)
Marking scheme: Solar cells / photovoltaic panels (1); light sensors, automatic doors, burglar alarms, light meters (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Photoelectric effect: stopping voltage and Planck’s constant
Aim: To investigate the photoelectric effect, to show that the maximum kinetic energy of the electrons depends on the frequency and not on the intensity of the light, and to determine Planck’s constant and the work function of a metal.
Apparatus (real laboratory): photocell, light source with filters, variable DC supply, voltmeter, microammeter.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Caesium and a wavelength of 400 nm. Read the current with the anode voltage at 0 V.
- Make the voltage negative, slowly (0.01 V steps), until the current just becomes 0.00 µA. Tap Record the stopping voltage.
- Repeat for 450, 500, 550 and 580 nm. Look at the graph of stopping voltage against frequency.
- Find h = e × gradient and the threshold frequency from the intercept on the frequency axis.
- Drag the lamp nearer: what happens to the current? Does the stopping voltage change?
- Choose Zinc. Try visible light, then ultraviolet (below 290 nm). Explain the result.
Results
| # | Metal | λ (nm) | f (× 10¹⁴ Hz) | Vs (V) | KEmax (eV) |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
Graph
Plot the stopping voltage Vs on the vertical axis against the frequency f. Find the gradient (h ÷ e) and the intercept on the frequency axis (threshold frequency f₀).
Questions
- What is the photoelectric effect? [2]
- What is the work function of a metal? [2]
- Why does red light not release electrons from zinc, however bright it is? [2]
- What happens to the current when the lamp is brought closer? Why? [2]
- Write Einstein’s photoelectric equation. [2]
- Calculate the energy of a photon of wavelength 500 nm in eV. [2]
- Light of frequency 7.5 × 10¹⁴ Hz falls on caesium (work function 2.1 eV). Calculate (a) the photon energy in eV, (b) the maximum kinetic energy of the electrons, (c) the stopping voltage. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C) [6]
- A student plots stopping voltage against frequency and obtains a straight line of gradient 4.1 × 10⁻¹⁵ V s, cutting the frequency axis at 5.1 × 10¹⁴ Hz. Find Planck’s constant and the work function in eV. [4]
- State three observations of the photoelectric effect that the wave theory of light cannot explain. [3]
- State two applications of the photoelectric effect. [2]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Photoelectric effect: stopping voltage and Planck’s constant
Expected results
Photon energy = 1240 ÷ λ(nm) eV. Work functions: caesium 2.10 eV (threshold 590 nm), potassium 2.30 eV (539 nm), calcium 2.90 eV (428 nm), zinc 4.30 eV (288 nm). Saturation current = 1.5 µA × (20 cm ÷ d)². For a negative anode voltage the current falls linearly to zero at Vs. Learners obtain h ≈ 6.6 × 10⁻³⁴ J s.
Answers
- What is the photoelectric effect? [2]
The emission of electrons from a metal surface when light (electromagnetic radiation) of high enough frequency falls on it.
- What is the work function of a metal? [2]
The minimum energy needed to remove an electron from the surface of the metal.
- Why does red light not release electrons from zinc, however bright it is? [2]
Each photon of red light has energy hf smaller than the work function of zinc; one photon gives its energy to one electron, so no electron can escape, whatever the number of photons.
- What happens to the current when the lamp is brought closer? Why? [2]
The current increases because more photons per second arrive, so more electrons are emitted per second; the stopping voltage does not change.
- Write Einstein’s photoelectric equation. [2]
hf = W + KEmax, where KEmax = eVs is the maximum kinetic energy of the emitted electrons.
- Calculate the energy of a photon of wavelength 500 nm in eV. [2]
E = hc ÷ λ = 6.63 × 10⁻³⁴ × 3.0 × 10⁸ ÷ 5.0 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J = 2.48 eV.
- Light of frequency 7.5 × 10¹⁴ Hz falls on caesium (work function 2.1 eV). Calculate (a) the photon energy in eV, (b) the maximum kinetic energy of the electrons, (c) the stopping voltage. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C) [6]
(a) E = hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J (1) = 3.1 eV (1). (b) KEmax = 3.1 − 2.1 (1) = 1.0 eV (1). (c) Vs = KEmax ÷ e = 1.0 V (2).
- A student plots stopping voltage against frequency and obtains a straight line of gradient 4.1 × 10⁻¹⁵ V s, cutting the frequency axis at 5.1 × 10¹⁴ Hz. Find Planck’s constant and the work function in eV. [4]
h = e × gradient (1) = 1.6 × 10⁻¹⁹ × 4.1 × 10⁻¹⁵ = 6.6 × 10⁻³⁴ J s (1). W = h f₀ (1) = 6.6 × 10⁻³⁴ × 5.1 × 10¹⁴ = 3.4 × 10⁻¹⁹ J = 2.1 eV (1).
- State three observations of the photoelectric effect that the wave theory of light cannot explain. [3]
No emission below a threshold frequency, however intense the light (1); KEmax depends on frequency, not intensity (1); emission is instantaneous even with very dim light (1).
- State two applications of the photoelectric effect. [2]
Solar cells / photovoltaic panels (1); light sensors, automatic doors, burglar alarms, light meters (1).