In short

Shine light on a clean metal surface. Above a threshold frequency f_0 (specific to the metal), electrons are ejected — the photoelectric effect. Below f_0, nothing happens, no matter how bright the light. Above f_0, the maximum kinetic energy of the ejected electrons depends on frequency, not intensity, and emission is instantaneous.

The wave theory of light predicts none of this. Einstein's 1905 explanation: light comes in discrete packets — photons — each carrying energy E = hf. A single photon either has enough energy to liberate an electron (if hf \geq \phi, where \phi is the metal's work function) or it does not. One photon, one electron. Energy balance gives Einstein's photoelectric equation:

\boxed{\;K_\text{max} = hf - \phi\;}

Planck's constant h = 6.626 \times 10^{-34} J·s. The threshold frequency is f_0 = \phi/h. The same equation tells you why a zinc plate emits electrons under ultraviolet but stays inert under the most intense red light — and why Einstein got the Nobel Prize for this, not for relativity.

Take a clean strip of zinc, connect it to a sensitive electrometer, and shine a desk lamp on it. Nothing happens. Swap the desk lamp for a mercury vapour lamp that emits ultraviolet. The electrometer needle swings immediately — electrons are leaving the zinc. Now dim the ultraviolet lamp until it is a hundred times fainter. The needle still swings, just more slowly. Turn the desk lamp back on at full brightness. Still nothing.

Something is wrong with the intuition that "more light means more energy." A faint source of one colour can knock electrons out of a metal that a brilliant source of another colour cannot touch. Light is doing something to the electrons that does not scale with how bright the light is — only with which light it is.

The puzzle is deeper than it looks. Before 1905, every physicist knew that light was a wave. Young's double-slit experiment, Maxwell's equations, the entire machinery of electromagnetism — all of it said light was a continuous oscillation of electric and magnetic fields. A brighter wave carries more energy. Give a metal enough time under a dim wave and eventually the electrons should soak up enough energy to escape.

That is not what happens. And the reason it does not happen is that light, at its most fundamental, is not a continuous wave. It is a stream of discrete energy packets, each of a fixed size set by the frequency. That is the claim that won Einstein the Nobel Prize in Physics in 1921.

The experiment — light on a metal surface

The setup is almost embarrassingly simple, which is part of why the result was so shocking.

Photoelectric effect apparatusEvacuated glass tube with a metal cathode on the left and a collector anode on the right. Light enters through a window and strikes the cathode. A battery biases the anode; a microammeter reads photocurrent. Electrons shown as small circles drift from cathode to anode.evacuated tubecathode (metal)work function φanodelight (frequency f)photoelectrons (e⁻)μAmicroammeter
An evacuated tube with a metal cathode on the left and a collector anode on the right. Light of frequency $f$ strikes the cathode; if the frequency is high enough, electrons are ejected and drift across the vacuum to the anode, producing a tiny current the microammeter can read.

Inside the evacuated tube, the cathode is a clean plate of the metal you are studying — sodium, potassium, caesium, zinc. Light enters through a quartz window (ordinary glass blocks ultraviolet). If the light has the right colour, electrons — called photoelectrons — are ejected from the cathode, cross the vacuum, and land on the anode. A microammeter in the external circuit reads the resulting current, called the photocurrent.

You can vary three knobs independently:

And you measure two quantities:

That last one is the most important. If you reverse the battery so the anode is negative with respect to the cathode, electrons now have to climb an electrical hill to reach the anode. Only the electrons whose kinetic energy exceeds eV can make it across. When you ramp up the reverse voltage until the photocurrent drops to exactly zero, you have found the voltage V_s that just barely stops the fastest electron the metal emitted. Energy conservation then tells you the maximum kinetic energy of the emitted electrons:

K_\text{max} = e V_s

Why: an electron of charge e climbing a potential difference V_s loses energy eV_s. The fastest electron just barely makes it, which means its entire kinetic energy converted to electrical potential energy. Measure V_s, multiply by the electron charge, and you have read off K_\text{max} directly.

This is the experimental handle that lets you extract a clean number — K_\text{max} in joules — from a tabletop apparatus.

The observations — four stubborn facts

When Philipp Lenard did this experiment carefully in the 1890s and early 1900s, four facts emerged. Each one contradicts the wave theory of light.

Fact 1: There is a threshold frequency

For every metal, there is a threshold frequency f_0 below which no photoelectrons are emitted at all, no matter how intense the light or how long you wait. Shine the brightest red laser you can buy on a sheet of zinc, and zero electrons come off — forever. Shine a faint ultraviolet beam on the same zinc and electrons come off immediately. The cutoff is sharp. For zinc, f_0 corresponds to a wavelength of about 290 nm, well into the ultraviolet; for caesium, it corresponds to about 660 nm, deep in the visible red — which is why caesium cells are used in old photomultiplier tubes and solar-triggered street lights.

Fact 2: K_\text{max} depends on frequency, not intensity

Above the threshold, if you increase the frequency, the maximum kinetic energy of the photoelectrons increases. But if you hold the frequency fixed and increase the intensity, K_\text{max} does not budge. More intense light releases more electrons (the photocurrent goes up), but each electron leaves with the same energy distribution.

A graph of stopping voltage V_s versus frequency f is a straight line. Below f_0 there is no photocurrent at all, so V_s is undefined; above f_0, the line climbs with a slope that is the same for every metal — it depends only on a universal constant.

Fact 3: Emission is instantaneous

Turn the light on, and electrons come off immediately — within nanoseconds, which was already "instantaneous" by the standards of 1905 instruments, and we now know is within 10^{-9} s. There is no delay, no build-up, no lag. Even at intensities so low that a single electron's worth of energy should take minutes to accumulate on a classical wave picture, emission is still immediate.

Fact 4: Photocurrent saturates with voltage

At zero bias, some photoelectrons reach the anode. As you increase the forward voltage (anode positive), more are pulled across, and the photocurrent rises. But at some voltage, the current levels off at a saturation value that depends only on the light intensity — not on the voltage. Every single electron ejected from the cathode is now reaching the anode; you cannot collect more than the light released.

Why the wave theory cannot explain this

Here is the wave-theory prediction, sentence by sentence, against the experimental reality.

Wave theory says: energy is spread continuously across the wavefront. A brighter wave delivers more energy per unit area per second. Given enough time, any wave — of any frequency — can pump enough energy into an electron to knock it free.

Experiment says: below f_0, no electron is ever ejected, no matter how bright or how long.

Wave theory says: a brighter wave should give more energetic electrons. The electrons absorbing the wave have more energy available.

Experiment says: brighter light gives more electrons, each with the same maximum energy. Intensity does not affect K_\text{max}.

Wave theory says: at low intensity, it should take a measurable time for an electron to absorb enough energy from a continuous wave to escape. For a faint source, this delay could be minutes.

Experiment says: emission is immediate at any intensity.

Three direct contradictions, from four clean observations. The wave theory of light — triumphant for a century after Young, solidified by Maxwell, seemingly unassailable — cannot describe what happens when light meets a metal surface.

Einstein's 1905 explanation — light as photons

In 1905 — the same year he published special relativity — Einstein proposed something audacious. What if light is not a continuous wave but a stream of discrete packets, each carrying a fixed amount of energy determined by the frequency? He borrowed the idea from Planck's 1900 analysis of blackbody radiation, but he made it physical: Planck had said the oscillators in a hot body exchange energy in quanta; Einstein said the light itself travels in quanta.

Each packet — now called a photon — carries energy

\boxed{\;E = hf\;}

where h = 6.626 \times 10^{-34} J·s is Planck's constant and f is the frequency of the light. A brighter beam of the same colour has more photons per second; it does not have more energetic photons. The colour sets the energy per photon; the intensity sets the number of photons per second.

With that single move, every observation falls into place.

Why a threshold frequency? Each electron inside the metal is held in by an energy barrier called the work function \phi — the minimum energy needed to yank it out. Ejection is a one-photon-one-electron event: a single photon either delivers enough energy (hf \geq \phi) to break the electron free, or it does not. Below f_0 = \phi/h, no single photon has enough punch. Adding more photons does not help — each electron interacts with one photon at a time. The threshold is sharp because it is the atomic-scale energy budget that matters, not the total wave energy.

Why does K_\text{max} depend on frequency, not intensity? The photon gives its energy hf to the electron. Some of that energy (exactly \phi) is spent breaking the electron out of the metal. Whatever is left shows up as kinetic energy:

K_\text{max} = hf - \phi

Higher frequency means more energy per photon, so more energy left over after paying \phi. Increasing intensity means more photons per second — hence more electrons per second (the photocurrent goes up) — but each individual ejection event is governed by the same hf - \phi balance. K_\text{max} is a single-photon property.

Why is emission instantaneous? A photon either hits an electron or it does not. When it does, the whole transaction happens in one step — there is nothing to accumulate, no build-up. Emission is instantaneous by construction.

Why does photocurrent saturate with voltage? Once the forward voltage is high enough that every emitted electron reaches the anode, increasing the voltage further cannot make a difference — you are already collecting all of them. The saturation level is set by the rate at which photons arrive and eject electrons — the light intensity — not by the voltage.

Einstein's equation is a simple statement of energy conservation, applied one photon at a time:

\boxed{\;K_\text{max} = hf - \phi\;}

Rewriting in terms of stopping voltage:

eV_s = hf - \phi
V_s = \frac{h}{e}\,f - \frac{\phi}{e}

Why: substitute K_\text{max} = eV_s on the left, then divide through by e. This is now the equation of a straight line: V_s as a function of f, with slope h/e and intercept -\phi/e. The slope h/e is a universal constant — it must come out the same for every metal. Only the intercept (and hence the threshold f_0 = \phi/h) depends on the metal. This is Millikan's 1916 test, and it is how h was measured to high precision.

The classic K_\text{max} vs f graph

If the photon hypothesis is right, a plot of K_\text{max} (or equivalently eV_s) against frequency f should be a straight line whose slope is Planck's constant — same slope for every metal, with only the intercept shifting. This is the graph that settled the matter experimentally in 1916, and it is the graph that JEE loves to set problems on.

The interactive below lets you dial the frequency of the incoming light and watch the photoelectric equation play out live. The vertical dashed line is the threshold frequency f_0; below it, no electrons come off.

Interactive: maximum kinetic energy of photoelectrons as a function of frequency Graph of K_max in electron-volts versus light frequency in hertz for sodium. The line is K_max equals (h over e) times f minus phi over e. Below threshold frequency 5.5 times ten to the fourteen hertz, K_max is negative (no emission). Drag the red point to change the frequency and see the value of K_max read out live. frequency f (×10¹⁴ Hz) K_max (eV) 0 1 2 3 4 5 10 15 K_max = 0 threshold f₀ 5.5 K_max = hf − φ (sodium, φ = 2.28 eV) drag the red point along the f-axis
Drag the red point along the horizontal axis to change the frequency of the incoming light on a sodium surface ($\phi = 2.28$ eV). The red curve is the photoelectric equation $K_\text{max} = hf - \phi$. Below the threshold $f_0 \approx 5.5 \times 10^{14}$ Hz (dashed vertical), $K_\text{max}$ would come out negative — meaning no photon has enough energy, so nothing is emitted. Above threshold, every extra hertz of frequency buys exactly $h = 4.136 \times 10^{-15}$ eV·s more kinetic energy.

A few things to read off that graph by eye:

Worked examples

Example 1: Sodium under green light

Sodium metal has a work function \phi = 2.28 eV. Green light from a mercury lamp has wavelength \lambda = 546 nm. Will sodium emit photoelectrons, and if so, what is the maximum kinetic energy of the emitted electrons? What stopping voltage would just halt them?

Energy balance for sodium under 546 nm green lightA horizontal bar labelled hf equals 2.27 eV, divided into two parts. The left part (2.28 eV) is labelled phi (work function). The right part (a thin sliver) would be K_max but is zero because hf is just below phi.Photon energy hf = 2.27 eV↑ needed: φ = 2.28 eV (work function)hf < φ, so no emission(shortfall: 0.01 eV — photon is just below threshold)
The 546 nm green photon carries 2.27 eV — almost, but not quite, enough to pay sodium's 2.28 eV work function. The electron cannot escape.

Step 1. Convert wavelength to frequency and photon energy.

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{546 \times 10^{-9} \text{ m}} = 5.494 \times 10^{14} \text{ Hz}
E_\text{photon} = hf = (4.136 \times 10^{-15} \text{ eV·s})(5.494 \times 10^{14} \text{ Hz}) = 2.272 \text{ eV}

Why: for photon energies in eV, using h in eV·s (4.136 \times 10^{-15}) skips the conversion through joules and elementary charge. A useful shortcut worth memorising: hc/e \approx 1240 eV·nm, so E_\text{photon}\text{ (eV)} = 1240/\lambda\text{ (nm)}. Check: 1240/546 = 2.271 eV — the same answer.

Step 2. Compare to the work function.

E_\text{photon} = 2.272 \text{ eV} < \phi = 2.28 \text{ eV}

Why: Einstein's equation gives K_\text{max} = hf - \phi = -0.008 eV, a negative number. Kinetic energy cannot be negative — the photon simply does not have enough energy to free the electron. The interaction ends with the photon being absorbed or re-emitted elsewhere, and the electron staying put.

Step 3. Conclusion.

Result: No photoelectron emission. Sodium is (just barely) opaque to 546 nm green light at the photoelectric level — you would need a slightly shorter wavelength (say 540 nm or below) to start seeing photocurrent.

What this shows: The threshold is sharp and unforgiving. A wavelength just barely too long means zero photoelectrons; a wavelength just barely short enough means every photon can potentially eject an electron. In practice sodium cells are used with blue and violet light (400–500 nm) to stay comfortably above threshold.

Example 2: Caesium under ultraviolet — measuring Planck's constant

Caesium has a work function \phi = 2.14 eV. You shine ultraviolet light of wavelength \lambda = 200 nm on a clean caesium cathode. Find (a) the maximum kinetic energy of photoelectrons, (b) the stopping voltage, (c) the maximum speed of the photoelectrons. Assume m_e = 9.11 \times 10^{-31} kg.

Energy balance for caesium under 200 nm ultraviolet lightA horizontal bar representing photon energy 6.20 eV. The left 2.14 eV is labelled phi (work function). The right 4.06 eV is labelled K_max.Photon energy hf = 6.20 eVφ = 2.14 eVK_max = 4.06 eVtotal energy hf = 6.20 eV
Photon energy 6.20 eV splits into the work function (2.14 eV, grey) and the electron's kinetic energy (4.06 eV, red). This bar-chart picture is Einstein's equation drawn to scale.

Step 1. Photon energy.

E_\text{photon} = \frac{1240 \text{ eV·nm}}{200 \text{ nm}} = 6.20 \text{ eV}

Why: the shortcut E_\text{photon}\text{ (eV)} = 1240/\lambda\text{ (nm)} is doing hc/\lambda in one step.

Step 2. Maximum kinetic energy.

K_\text{max} = hf - \phi = 6.20 - 2.14 = 4.06 \text{ eV}

Why: Einstein's equation directly. The photon pays the work function toll and the leftover becomes kinetic energy.

Step 3. Stopping voltage.

V_s = \frac{K_\text{max}}{e} = 4.06 \text{ V}

Why: when K_\text{max} is in electron-volts, V_s in volts is numerically equal. That is the very convenience of the eV unit.

Step 4. Maximum speed.

Convert K_\text{max} to joules: 4.06 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 6.506 \times 10^{-19} J.

K_\text{max} = \tfrac{1}{2} m_e v_\text{max}^2 \quad\Rightarrow\quad v_\text{max} = \sqrt{\frac{2 K_\text{max}}{m_e}}
v_\text{max} = \sqrt{\frac{2 \times 6.506 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{1.429 \times 10^{12}} \approx 1.195 \times 10^6 \text{ m/s}

Why: the kinetic energy formula K = \frac{1}{2}mv^2 inverted. The speed is about 0.4% of the speed of light — fast, but comfortably non-relativistic, so the classical formula is fine. (If the photon energy exceeded about 10 keV, you would need the relativistic kinetic-energy formula.)

Step 5. Verify the units.

\sqrt{\text{J/kg}} = \sqrt{\text{kg·m}^2/\text{s}^2 / \text{kg}} = \sqrt{\text{m}^2/\text{s}^2} = \text{m/s}. Good.

Result: K_\text{max} = 4.06 eV, stopping voltage V_s = 4.06 V, v_\text{max} \approx 1.2 \times 10^6 m/s.

What this shows: Every photoelectric problem at JEE level is this exact sequence: find the photon energy from \lambda, subtract the work function to get K_\text{max}, and (if asked) compute the stopping voltage or the electron speed. Memorising the 1240 eV·nm shortcut saves time on the exam.

Example 3: Measuring Planck's constant from a $V_s$–$f$ graph

A student shines light of three different frequencies on a potassium cathode and measures the stopping voltage each time:

f (×10¹⁴ Hz) V_s (V)
6.0 0.236
8.0 1.064
10.0 1.892

From this data, estimate Planck's constant h and the work function \phi of potassium.

The three data points lie almost exactly on the fitted line $V_s = (h/e)f - \phi/e$. The horizontal intercept $f_0 \approx 5.4 \times 10^{14}$ Hz is the threshold frequency; the slope $h/e$ extracted from the line gives Planck's constant directly.

Step 1. Write the predicted relationship.

V_s = \frac{h}{e}\,f - \frac{\phi}{e}

The slope of a V_s vs f plot is h/e; the vertical intercept is -\phi/e.

Why: this is Einstein's equation divided by e. If the photon hypothesis is right, the data must lie on a straight line whose slope is a universal constant.

Step 2. Compute the slope from two of the data points.

\text{slope} = \frac{V_s(2) - V_s(1)}{f(2) - f(1)} = \frac{1.064 - 0.236}{(8.0 - 6.0) \times 10^{14}} = \frac{0.828}{2.0 \times 10^{14}} = 4.14 \times 10^{-15} \text{ V·s}

Why: pick any two points and compute rise-over-run. A good check: use the other pair. (1.892 - 0.236)/(4.0 \times 10^{14}) = 4.14 \times 10^{-15}. The slope is consistent across pairs — the data is linear, as the photon hypothesis predicts.

Step 3. Extract Planck's constant.

h = e \times \text{slope} = (1.602 \times 10^{-19} \text{ C})(4.14 \times 10^{-15} \text{ V·s}) = 6.63 \times 10^{-34} \text{ J·s}

Why: multiplying slope by e converts from eV·s to J·s. The accepted value is 6.626 \times 10^{-34} J·s — the student is within 0.1%, which is excellent for a tabletop lab.

Step 4. Extract the work function from the intercept.

At the threshold V_s = 0, so f_0 = \phi e / (h) \cdot \text{intercept} — or more directly, extrapolate the line: V_s = 0 when f = f_0. From the slope and one point:

f_0 = f_1 - \frac{V_s(1)}{\text{slope}} = 6.0 \times 10^{14} - \frac{0.236}{4.14 \times 10^{-15}} = 6.0 \times 10^{14} - 0.57 \times 10^{14} = 5.43 \times 10^{14} \text{ Hz}
\phi = h f_0 = (4.14 \times 10^{-15} \text{ eV·s})(5.43 \times 10^{14} \text{ Hz}) = 2.25 \text{ eV}

Why: the threshold frequency is where the line crosses zero. Multiply by h (in eV·s so \phi comes out in eV directly) to get the work function. Potassium's accepted work function is about 2.24 eV — again, essentially exact.

Result: h \approx 6.63 \times 10^{-34} J·s (accepted: 6.626 × 10⁻³⁴), \phi \approx 2.25 eV.

What this shows: This is exactly how Robert Millikan confirmed Einstein's photon hypothesis in 1916 — ten years after the prediction, using painstakingly clean metal surfaces. The beauty of the experiment is that the slope comes out to Planck's constant regardless of which metal you use. A table-top apparatus run by a careful student measures one of the most fundamental constants in physics to within a percent. This is the kind of experiment that is done in Indian college labs at BARC's training programs and at every IIT — the apparatus is standard, and the result is breath-taking.

Common confusions

Why this matters — ISRO rooftops, photodiodes, solar India

The photoelectric effect is not a laboratory curiosity. It is the physical basis of every technology that turns light into an electrical signal.

Every photodetector on Earth — and every photodetector in orbit around Earth — is a descendant of the zinc-plate experiment, reading out Einstein's equation one photon at a time.

If you came here to understand Einstein's equation, solve JEE problems on K_\text{max} and V_s, and know why solar cells exist, you have what you need. What follows is for readers who want to see why the photon hypothesis runs deeper than the photoelectric effect alone, and to meet the historical loose ends.

The wave theory's failure in numbers

How bad was the classical prediction? Consider a faint UV source — say, a beam of intensity I = 10^{-10} W/m² (about a nanowatt per square centimetre). A metal electron has an effective "target area" of about \pi r_a^2 where r_a \sim 10^{-10} m (the atomic radius), so the classical wave power absorbed by one electron is roughly

P_\text{absorbed} \sim I \times \pi r_a^2 = 10^{-10} \times 3 \times 10^{-20} = 3 \times 10^{-30} \text{ W}

To accumulate 2 eV = 3.2 \times 10^{-19} J of energy:

\Delta t \sim \frac{3.2 \times 10^{-19}}{3 \times 10^{-30}} \approx 10^{11} \text{ s} \approx 3000 \text{ years}

The classical theory predicts a three-thousand-year delay before the first electron is emitted. The experiment sees emission in nanoseconds. This is not a small discrepancy — it is twenty orders of magnitude, and it kills the wave theory for this process outright.

The photon and the full wave description coexist

Einstein did not kill the wave picture of light — he augmented it. A light beam is both a wave (it interferes, it diffracts, it polarises) and a stream of photons (it ejects electrons one at a time, it is absorbed in discrete quanta). The same beam shows wave behaviour in a Young's double-slit experiment and particle behaviour in a photoelectric cell. This is wave-particle duality, and it is not a contradiction — it is a statement that light is neither a classical wave nor a classical particle but a quantum-mechanical object whose classical limits are the two pictures. The article on photons and wave-particle duality develops this in detail.

The momentum of a photon

A photon carries not just energy E = hf but also momentum

p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda}

This momentum is what drives radiation pressure — sunlight pushing on a comet's dust tail, or on the sails of a hypothetical light-sail spacecraft. The relation p = h/\lambda is also the seed of the de Broglie hypothesis: if a photon with wavelength \lambda has momentum h/\lambda, maybe a particle with momentum p has a wavelength h/p too. That turned out to be exactly right. The article on the de Broglie hypothesis follows that thread.

The Compton effect — photons as billiard balls

Einstein's photon picture was fully vindicated in 1923 by a different experiment: Arthur Compton scattered X-rays off loosely bound electrons and observed that the scattered X-rays had a longer wavelength than the incident ones, with the shift depending on scattering angle in exactly the way a billiard-ball collision between a photon (momentum h/\lambda, energy hf) and a stationary electron would predict. The photon behaves like a particle with momentum, not just an energy packet. By the mid-1920s, the photon had stopped being Einstein's bold conjecture and become standard physics.

Why did it take until 1921 for the Nobel?

Einstein's 1905 paper was regarded as radical bordering on heretical. Even Planck, whose own quantum hypothesis had inspired it, thought Einstein had gone too far. It took Millikan's meticulous 1916 experimental confirmation — Millikan himself had initially set out to disprove Einstein — to convince the community. When the Nobel Committee awarded Einstein the 1921 Physics Prize, the citation carefully singled out "his discovery of the law of the photoelectric effect" rather than relativity, which was still considered controversial. Einstein gave the prize money (and more) to his ex-wife Mileva Marić as part of their divorce settlement.

Saha ionisation — India's contribution to photon-driven physics

Meghnad Saha's 1920 ionisation equation — which predicts how many atoms in a hot gas are ionised by the photons pouring out of a stellar atmosphere — is the astrophysical counterpart of the photoelectric effect, using exactly the same "one photon, one bound electron" logic applied to atoms in stars instead of electrons in metals. The Saha equation is the reason astronomers can read temperatures off stellar spectra. It is a direct descendant of Einstein's photon, and it is a centrepiece of Indian theoretical physics.

Where this leads next