Why can a dim blue light free electrons from a metal when the brightest red light cannot?

How to use this simulation

1. Start in the Vacuum tube view. The plate is sodium (φ = 2.3 eV) and the light is 400 nm. Each photon carries 3.10 eV, more than the work function, so electrons (blue dots) leave the plate. The reading says each has at most 0.80 eV.
2. Drag the wavelength up toward red. Around 539 nm the electrons stop coming out and a message appears. Slide it back and they return. Now drag the intensity to 100 %: more photons arrive, but the electrons set off no faster.
3. Change the plate to zinc and then copper. With the same 400 nm light nothing happens, because 400 nm photons are too weak for those metals. Go to short wavelengths and electrons appear again.
4. Slide the collector voltage negative. This is a retarding voltage: the electrons with the least energy turn around first, and the current falls. At the stopping potential V₀ the strongest electron is turned back as well, and the current is zero. Tick "Snap to the stopping potential" to land on it, and read V₀ in the readings: it equals K in volts.
5. Switch to the K against f graph. Each metal gives a straight line that starts on the frequency axis at its threshold. The three lines are parallel, because they all have the same slope, h/e. The dashed marker follows your light.
6. Switch to the I against V graph. The dashed curve is another intensity of the same light. It is taller or shorter, but it hits the voltage axis at exactly the same stopping potential.

The puzzle that waves could not solve

In the late 1800s, physicists found that light shining on a metal could knock electrons out of it. The effect was easy to measure: put the metal plate in an evacuated tube, collect the electrons on a second plate and watch an ammeter. Light as a wave seemed to make sense of this. A wave carries energy, the electrons absorb some, and when they have enough, they escape.

But the measurements disagreed with that picture in three ways. First, below a certain frequency no electrons came out at all, however bright the light and however long you waited. Second, once the frequency was above that threshold, the electrons appeared at once, with no delay, even in extremely dim light. A wave spreading over many atoms should have needed time to build up enough energy in one electron. Third, the energy of the electrons depended on the frequency of the light and not on its brightness. A wave theory says a brighter wave should hurl electrons out harder.

Try the first and third observations in the animation. Drag the intensity: the current changes, the maximum electron energy does not. Drag the wavelength: the electron energy changes at once and below the threshold there are no electrons whatever the brightness.

Einstein’s answer: light comes in photons

In 1905 Einstein proposed that light is made of packets of energy, called photons. Each photon of frequency f carries energy E = hf, where h = 6.626 × 10⁻³⁴ J·s is Planck’s constant. A photon gives all of its energy to a single electron, or none of it. There is no building up energy over time and no sharing between electrons.

Wavelength is often easier to read, so use E = hc/λ. With h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s and 1 eV = 1.602 × 10⁻¹⁹ J, this becomes E (in eV) = 1,241 / λ (in nm). The 400 nm light of the animation has f = 7.50 × 10¹⁴ Hz and photon energy 3.10 eV, which is 4.97 × 10⁻¹⁹ J.

Brightness now has a clear meaning. Intensity is the number of photons arriving each second. Doubling it doubles the number of photons, so it doubles the number of electrons, but each photon still carries the same hf. That is how the three puzzles disappear. A dim blue lamp sends few photons, each strong enough. A bright red lamp sends a flood, every one of them too weak.

Work function and Einstein’s equation

Electrons are held inside a metal. To escape, an electron must be given a minimum energy, called the work function φ of the metal. It depends on the metal; the animation uses typical textbook values, which move a little with the state of the surface. Energy that is left after paying φ becomes the electron’s kinetic energy.

That gives Einstein’s photoelectric equation: hf = φ + K_max. K_max is the greatest kinetic energy because an electron deep in the metal spends more on its escape and leaves slower; only those near the surface have the maximum. In the animation the electrons’ energies are spread between zero and K_max.

For sodium (φ = 2.3 eV) in 400 nm light, K_max = 3.10 − 2.3 = 0.80 eV. The fastest electrons move at about 5.3 × 10⁵ m/s, from K = ½mv² with m = 9.109 × 10⁻³¹ kg.

The threshold: no electrons below f₀

The smallest photon that can free an electron has hf = φ, so K_max = 0. The frequency where this happens is the threshold frequency f₀ = φ / h, and the matching wavelength λ₀ = hc / φ. For sodium, f₀ = 5.56 × 10¹⁴ Hz and λ₀ = 539 nm. Light with a longer wavelength or lower frequency than that does nothing, however bright.

Because the threshold is set by the metal, different metals respond to different parts of the spectrum. Sodium works with visible light all the way up to green. Zinc and copper need ultraviolet. The table puts all three in the same lights. A dash means that the photon energy is below φ and no electron is released.

Intensity changes the current, frequency changes the energy

Each photon above the threshold frees at most one electron, so the photocurrent is proportional to the number of photons per second, that is, to the intensity. In the animation a lamp at 100 % gives a saturation current of 2.0 µA, so the current at 60 % is 1.20 µA. The saturation current is the most the tube can deliver: every electron that leaves is collected.

Frequency does the opposite job. Increase it and each photon carries more energy, so K_max rises in a straight line, but the number of electrons per second is unchanged. The two controls are independent, and that independence is the whole point of the experiment.

Do not confuse the two sides of an exam question. “The light is made twice as bright” changes the current and not K_max or V₀. “The frequency is doubled” changes K_max and V₀ (not by a factor of two, because φ stays the same) and leaves the number of photons alone.

The stopping potential

Now give the collector a negative voltage with respect to the plate. It pushes the electrons back. An electron with less than e|V| of kinetic energy turns around before it arrives, so the current falls as the voltage becomes more negative. When the voltage reaches −V₀ even the fastest electron is turned back and the current is zero. This is the stopping potential: eV₀ = K_max.

The animation spreads the electron energies evenly from zero to K_max, so the current falls along a straight line. With 0.5 V of retarding voltage on the default setup the current has dropped from 1.20 µA to 0.45 µA. Real curves round off near the ends, but the stopping potential, where the current reaches zero, is the same.

The stopping potential for the sodium default is 0.80 V, equal to K_max in electron-volts. That makes V₀ a convenient, directly measurable stand-in for K_max, and it is how experimenters weighed the energy of photoelectrons without measuring their speed.

The K against f graph and Planck’s constant

Rearrange Einstein’s equation as K_max = hf − φ. It is a straight line in f. Its slope is h (or h/e, if you plot V₀ in volts instead), its x-intercept is the threshold frequency f₀ and its y-intercept is −φ. Every metal gives a line with the same slope, so the lines for different metals are parallel and only shift sideways.

The slope in the K graph above is h/e = 4.136 × 10⁻¹⁵ V·s. Millikan’s careful experiments in 1916 measured this slope and found Planck’s constant, confirming Einstein’s equation even though Millikan had set out hoping to disprove the photon idea. This is why the graph is so often asked about.

Solved problems

Each answer uses the same functions as the animation, so you can check any of them by setting the sliders.

Mistakes that cost marks

Check this list before the exam. Each one has cost someone a mark.

• Saying a brighter light gives faster electrons. It gives more electrons per second, not faster ones.
• Saying that a very bright light of low frequency will eventually work. Below f₀ it never will, because each photon is alone.
• Forgetting to subtract the work function. K_max = hf − φ, not hf.
• Mixing units. If φ is in eV, convert the photon energy to eV (E = 1240 / λ) before subtracting, or convert φ to joules.
• Treating the stopping potential as a property of the intensity. It depends on frequency and metal only.
• Reading the K against f graph wrongly: the slope is h, the x-intercept is f₀, the y-intercept is −φ, and the line does not pass through the origin.
• Using the wave picture to explain the delay: there is none. The emission is effectively instantaneous.

Where else you meet it

Light meters in cameras, automatic doors, smoke detectors and the light sensor that turns a street lamp on at dusk all use the effect or its relatives. Solar cells use a cousin of it inside a semiconductor, where photons lift electrons across a gap. Night-vision tubes turn faint photons into electrons and multiply them. And the idea that light comes in packets was one of the first steps toward quantum mechanics.

If you want to see the wave side of light, the interference, diffraction and polarisation pages show behaviour that photons alone cannot explain. Light needs both pictures, which is the dual nature of radiation.

Frequently asked questions