Photoelectric Effect Simulator

What the photoelectric effect is

The photoelectric effect is the release of electrons from a metal by light, and this simulation shows its two rules at work: below the metal’s threshold frequency no electron escapes however bright the light, and above it the fastest electron leaves with K_max = hf − φ. Here hf is the energy of one photon of frequency f and φ is the work function, the least energy that frees an electron from that surface. Each electron takes in a single photon, so the threshold is f₀ = φ/h.

The stopping potential, also called the cut-off voltage, measures what is left. Make the anode negative relative to the metal and it pushes the electrons back; at V₀ = K_max/e not even the fastest reaches it and the current stops. An energy in electronvolts and a voltage in volts are the same number, so an electron that leaves with 0.74 eV is stopped by 0.74 V. The Photon Energy Calculator turns any wavelength into electronvolts on its own.

How to use the simulation

Set the light by wavelength or by frequency, choose the intensity and the metal, then move the anode voltage and watch the current. The readouts give the stopping potential, the maximum kinetic energy, the photon energy, the work function, the threshold frequency and wavelength, the fastest electron’s speed and the current reaching the anode. The Wavelength and Frequency Calculator converts between the two ways of describing the light.

• Intensity counts photons per second. It changes how many electrons leave, and so the size of the current, but never their energy.
• Metal sets the work function from Kaye and Laby’s table, and Custom takes any value from 1.20 eV to 6.00 eV, so a textbook problem can be set up exactly.
• Anode voltage runs from −5 V, which holds electrons back, to +2 V, which pulls every one across. Every stopping potential the settings can produce lies inside that range.
• The current against voltage plot shows the current falling to zero at minus the stopping potential, with the full-intensity curve dashed behind it whenever the intensity is lower.
• The stopping potential against frequency plot is a straight line of slope h/e, with a dot on the light you have set.
• The picture frees one electron for every photon so that the exchange can be seen. Coloured electrons are heading for the anode and grey ones have been turned back. The bar on the right is Einstein’s equation drawn to scale: the photon’s energy hf, the part φ spent escaping and the part K left over.

Worked example: violet light on sodium

What does 400 nm light do to sodium, whose work function Kaye and Laby give as 2.36 eV? These are the simulation’s opening settings.

• Photon energy: E = hc/λ = 1239.84 eV·nm ÷ 400 nm = 3.0996 eV, at a frequency of c/λ = 749.5 THz.
• Maximum kinetic energy: K_max = 3.0996 − 2.36 = 0.7396 eV, which is 1.185 × 10⁻¹⁹ J.
• Stopping potential: V₀ = K_max/e = 0.7396 V, so the anode must sit at −0.74 V to stop the current.
• Threshold: f₀ = φ/h = 570.6 THz, a wavelength of 525.4 nm. Green light longer than that, and all yellow, orange and red light, frees nothing from sodium.
• Fastest electron: v = √(2K_max/m) = 5.10 × 10⁵ m/s, or 510 km/s.
• At −0.37 V, about half the stopping potential, the share of electrons that still get through is (1 − 0.37/0.7396)² = 0.25, a quarter, so at the opening intensity of 50 percent the relative current reads 0.125.

The readouts show the same figures to the precision the work function supports: 0.74 V, 570.6 THz, 525.4 nm and 510 km/s.

Worked example: caesium lit at 6 × 10¹⁴ Hz

An exercise in NCERT’s Class 12 physics textbook takes the work function of caesium as 2.14 eV and lights it at 6 × 10¹⁴ Hz. What are the maximum kinetic energy, the stopping potential and the fastest electron’s speed? Set the light by frequency to 600 THz and choose Custom at 2.14 eV.

• hf = 4.1357 × 10⁻¹⁵ eV·s × 6 × 10¹⁴ Hz = 2.4814 eV.
• K_max = 2.4814 − 2.14 = 0.3414 eV, so the stopping potential is 0.341 V.
• v = √(2 × 0.3414 eV × 1.602 × 10⁻¹⁹ J/eV ÷ 9.109 × 10⁻³¹ kg) = 3.47 × 10⁵ m/s.

NCERT’s worked examples take h = 6.63 × 10⁻³⁴ J·s and e = 1.6 × 10⁻¹⁹ C, which give 2.49 eV and 0.346 eV here and put caesium’s threshold at 5.16 × 10¹⁴ Hz, where the exact constants give 517.4 THz, or 5.174 × 10¹⁴ Hz. That is why a textbook answer and the simulation can differ in the third figure.

Why intensity changes the current but not the stopping potential

Intensity is the number of photons arriving each second, and each photon acts on one electron. More photons free more electrons and a larger current, but each electron still gets one photon’s worth of energy, so the fastest leaves with hf − φ at any brightness. Raise the intensity from 50 to 100 percent and the saturation current doubles while the current still falls to zero at the same voltage. Below the threshold even full intensity frees nothing.

A wave picture of light predicts the opposite on both counts: brighter light should give faster electrons, and any frequency should work if the light is bright enough for long enough. Neither happens, and emission starts within about 10⁻⁹ s even in dim light. Einstein’s 1905 proposal that light arrives in quanta of energy hf explains all three. Interference, which the Double Slit Simulator shows, needs light to behave as a wave; the photoelectric effect needs it to arrive in photons, and both are true.

Power is a different matter. At the same power, shorter wavelengths bring fewer photons, because each carries more energy: one watt of 400 nm light is 2.01 × 10¹⁸ photons a second and one watt of 600 nm light is 3.02 × 10¹⁸. The intensity control here counts photons, so the saturation current depends on it alone.

Stopping potential against frequency: a slope of h/e

Written as V₀ = (h/e)f − φ/e, Einstein’s equation says that a plot of stopping potential against frequency is a straight line with the same slope for every metal, h/e = 4.136 × 10⁻¹⁵ V·s, the Planck constant divided by the elementary charge. It crosses zero at the threshold frequency, and extended back to zero frequency it meets the voltage axis at −φ/e, so one line gives both Planck’s constant and the work function. The lower plot draws it for the chosen metal. Read the other way, a measured slope gives h: the 4.12 × 10⁻¹⁵ V·s of an NCERT exercise means h = 4.12 × 10⁻¹⁵ V·s × 1.602 × 10⁻¹⁹ C = 6.60 × 10⁻³⁴ J·s.

Millikan measured that line for sodium, and in 1916, using the known value of the electron’s charge, obtained a Planck constant close to the value found in an entirely different context, which confirmed Einstein’s equation. Einstein’s 1921 Nobel Prize in Physics was awarded for his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect.

Where the work functions come from

The metals listed use the photoelectric values in the National Physical Laboratory’s online edition of Kaye and Laby’s Tables of Physical and Chemical Constants, which selected them from reviews by Rivière (1969) and by Hölzl and Schulte (1979) and gives their typical error as 0.02 eV. A work function belongs to the surface as much as to the metal. The same table gives potassium 2.30 eV by the photoelectric method and 2.01 eV by contact potential difference, notes that different crystal faces differ and that contamination usually lowers the value, and so textbooks disagree. NCERT’s worked example gives caesium 2.14 eV where Kaye and Laby have 1.95 eV; choose Custom to use the figure a problem states. The table gives lithium, beryllium and magnesium no photoelectric value, so they are not listed.

Caesium, rubidium, potassium, sodium and barium respond to visible light, calcium only to violet light, and zinc and every metal below it in the table need ultraviolet.

What this simulation leaves out

• Temperature. The current is modelled at absolute zero, where it reaches zero exactly at the stopping potential along a parabola that touches the axis, DuBridge’s result of 1933. At room temperature a few electrons carry a little more than hf − φ, so a measured curve creeps towards the axis instead, and DuBridge put the resulting uncertainty in the stopping potential at several hundredths of a volt.
• Contact potential. Both plates are treated as having the metal’s work function. In a real tube the anode’s is different, and the contact potential difference between the two metals shifts the whole curve along the voltage axis.
• How many electrons each photon frees. The current is shown relative to the saturation current at full intensity because the model does not know it in amperes. In a real metal the number of electrons freed per photon depends on the metal and the frequency, and close to the threshold it falls smoothly to zero, as the Fowler equation quoted in the same section of Kaye and Laby describes; here the saturation current is the same at every frequency above the threshold.
• The shape of the tube. Flat parallel plates sort electrons by the energy they carry normal to the surface, which gives the parabola drawn here. A small emitter at the centre of a spherical collector sorts them by their total energy, and for that case DuBridge’s curve cuts the axis at a large angle instead. Textbook graphs also show the current still rising a little above zero volts before it saturates; here every emitted electron is collected from zero volts up.
• Two photons at once. Light intense enough for one electron to absorb two photons, as a focused pulsed laser can be, frees electrons below the threshold frequency. The model allows one photon per electron.
• Time. The picture is slowed enormously so that electrons can be followed: at the opening settings the fastest real electron crosses a centimetre in about 20 ns. Near the threshold the picture is also sped up, uniformly, so that the slowest electrons still move, and it says so.
• Light shorter than 200 nm. That is the vacuum ultraviolet, named because the oxygen in air absorbs it, and it is left out.

Common mistakes

• Expecting brighter light to give faster electrons. Intensity sets how many electrons leave. Only the frequency and the metal set how fast the fastest one moves.
• Leaving the wavelength in nanometres. With h and c in SI units, E = hc/λ needs λ in metres. Putting in 400 rather than 4 × 10⁻⁷ makes the energy 10⁹ times too small.
• Mixing electronvolts and joules. 1 eV is 1.602 × 10⁻¹⁹ J. K_max in electronvolts is the stopping potential in volts, but in joules it has to be divided by e first.
• Getting the threshold backwards. Light has to be above the threshold frequency, which means below the threshold wavelength. Wavelengths longer than λ₀ = hc/φ fail at any intensity.
• Assuming the same power means the same number of photons. At one watt, 400 nm light delivers fewer photons than 600 nm light, so with a lamp’s power held fixed the saturation current changes with the colour even above the threshold.
• Giving the stopping potential the wrong sign. The anode sits at −V₀ relative to the metal. The stopping potential is quoted as a positive number, the size of that retarding voltage.
• Expecting four-figure agreement with a textbook. Answers worked with h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s and e = 1.6 × 10⁻¹⁹ C drift in the third figure from ones worked with the exact constants, as the second example shows.