The Photoelectric Effect & Its Laws

The photoelectric effect stopped being a puzzle and became a tool. Every photomultiplier tube — the ultrasensitive light detector at the heart of medical PET scanners, particle-physics calorimeters, and old-school night-vision — begins with a photocathode ejecting electrons by exactly Hertz's effect, then multiplies that trickle into a measurable pulse. Photoelectron spectroscopy turns the effect into chemistry: shine ultraviolet or X-ray light of known frequency on a material, measure the kinetic energies of the electrons that come off, and you read the binding energies of the atoms directly, a technique that maps the electronic structure of everything from catalysts to superconductors. Solar cells and the light meters in every camera and phone rest on the same principle that a photon can promote an electron across an energy barrier. And the threshold law is not an abstraction: it is why welders' arc light and tanning beds emit ultraviolet that damages skin while brighter visible light does not — energy per photon, set by frequency, is what crosses the biological threshold, exactly as it crosses the metal's work function.

Picture a clean metal plate in an evacuated glass tube. Inside the metal, the outermost electrons are not bound to any one atom; they roam through the whole block like a gas — the same free-electron sea that makes metals conduct. What keeps them from simply leaking out into the vacuum is an energy barrier at the surface, a step they cannot climb without help. The smallest amount of energy that will lift the least tightly held electron over that step and free it is the metal's work function $\varphi$, measured in electronvolts: caesium asks only $2.14\ \mathrm{eV}$, sodium $2.28\ \mathrm{eV}$, tungsten a steep $4.5\ \mathrm{eV}$. Give an electron at least $\varphi$ and it escapes; give it more and it leaves with the surplus as kinetic energy. Shine light on that plate and, if the light is the right colour, electrons stream off. These ejected electrons are called photoelectrons — Hertz saw the effect in 1887, Hallwachs showed a charged zinc plate discharging under ultraviolet light in 1888, J. J. Thomson identified the flying particles as electrons in 1899, and Philipp Lenard turned it into precise measurement in 1902. The apparatus that made those measurements — a light-sensitive emitter, a collector to catch the electrons, a battery to push or oppose them, and a sensitive ammeter — is the photocell, and its behaviour is the whole of this article.

Now run the experiment. With the collector held positive, it sweeps up every electron the light frees, and the ammeter reads a steady photocurrent. Turn the collector negative instead — a retarding voltage — and you push back against the arriving electrons; only the fastest ones now reach the collector, and as you increase the retarding voltage the current falls, until at one special value, the stopping potential $V_0$, even the fastest electron is turned back and the current drops to zero. That stopping potential is a direct energy measurement: it tells you the maximum kinetic energy the light gave any electron. Vary the colour and brightness of the light while watching the saturation current and the stopping potential, and the four laws reveal themselves.

The laws are best read off the current–voltage graph of the photocell. The next figure shows two things at once: what happens when you brighten the light at fixed colour, and what happens when you change the colour.

Law 1 — the threshold frequency. For each metal there is a minimum frequency $f_0$ (equivalently a maximum wavelength $\lambda_0$) below which no electrons are emitted, however intense the light and however long it shines. Red light on caesium — or any visible light on zinc — can pour in for hours and eject nothing. Push the frequency above $f_0$ and emission begins at once. The threshold is a property of the metal:

where: $\varphi$ is the work function, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant, and $c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}}$. The physical reading is that a metal has a "colour toll": nothing bluer than a certain frequency gets in free. The quantum reason why the threshold energy is exactly $\varphi$ is Einstein's, developed in the next two articles; indeed eq (1) itself already contains Planck's $h$, so connecting a work function to a threshold frequency is the quantum result we quote here and establish next article — classically no such threshold exists at all. Here we read each metal's threshold value from $\varphi$ and take it as experimental fact. For caesium, using the handy shortcut $hc = 1240\ \mathrm{eV\,nm}$,

So caesium responds to anything from green-yellow ($579\ \mathrm{nm}$) on up into the blue and ultraviolet, but red light leaves it cold. Sodium ($\varphi = 2.28\ \mathrm{eV}$) has $\lambda_0 = 544\ \mathrm{nm}$; tungsten ($\varphi = 4.5\ \mathrm{eV}$) needs $\lambda_0 = 276\ \mathrm{nm}$, deep in the ultraviolet.

Law 2 — emission is instantaneous. However dim the light, the current appears within about $10^{-9}\ \mathrm{s}$ of switch-on. There is no lag while energy is "collected". This is the law that most brutally contradicts the wave picture, as the derivation below makes quantitative.

Law 3 — saturation current is proportional to intensity. At fixed frequency above threshold, brightening the light ejects more electrons per second. Once the collector voltage is high enough to sweep up every ejected electron, the current levels off at its saturation current, and

where: intensity is the light power per unit area. Double the brightness and you double the number of photoelectrons per second, hence double the saturation current — but, crucially, not their energy. Intensity controls how many, never how fast.

Law 4 — maximum kinetic energy depends on frequency, not intensity. The stopping potential $V_0$ measures the maximum kinetic energy directly, because turning back the fastest electron costs exactly its kinetic energy:

where: $e = 1.60\times10^{-19}\ \mathrm{C}$ is the electron charge and $V_0$ the stopping potential in volts. The experimental finding is that $V_0$ — and therefore $KE_{\max}$ — rises linearly with the light's frequency and does not change when you brighten the light. Make the beam a hundred times brighter at the same colour and the stopping potential does not budge; nudge the frequency up and $V_0$ climbs immediately. The slope of the $V_0$-versus-$f$ line turns out to be the universal constant $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$ — a result we merely quote here and prove in Einstein's equation.

A classical light wave carries energy smoothly and continuously, spread over its whole wavefront, and delivers more of it when it is brighter. That picture makes three clear predictions, and experiment demolishes all three. First, there should be no threshold: energy is energy, so a dim blue beam and a blazing red one that deliver the same power should behave alike — turn up any colour bright enough and electrons must eventually boil off. They do not; below $f_0$, nothing happens at any brightness. Second, a very dim wave should show a time lag: an electron sitting in a small patch of the surface collects energy from the wave drop by drop, and if the wave is faint it should take a measurable while to accumulate the escape energy $\varphi$. Third, brighter light means a stronger oscillating electric field, which should shake electrons harder and send them off faster — so $KE_{\max}$ ought to grow with intensity. Instead $KE_{\max}$ ignores intensity entirely and tracks frequency. The one thing classical theory gets right is Law 3, that brighter light means more current — and that is precisely because "more energy per second" can free "more electrons per second". The derivation below makes the time-lag failure numerical, because a number is more convincing than a paragraph.

Board level — thresholds of sodium (CBSE and ICSE pattern)

Sodium has a work function $\varphi = 2.28\ \mathrm{eV}$. Take $h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}}$, $e = 1.60\times10^{-19}\ \mathrm{C}$, and $hc = 1240\ \mathrm{eV\,nm}$. Find (a) the threshold wavelength, (b) the threshold frequency, and decide whether (c) blue light of $480\ \mathrm{nm}$ and (d) the yellow sodium line at $589\ \mathrm{nm}$ can eject photoelectrons.

Step 1. Threshold wavelength: $\lambda_0 = hc/\varphi = 1240/2.28 = \boxed{544\ \mathrm{nm}}$. Any wavelength shorter than this (higher frequency) can eject electrons; anything longer cannot.

Step 2. Threshold frequency: $f_0 = \varphi/h = (2.28\times1.60\times10^{-19})/(6.63\times10^{-34}) = \boxed{5.50\times10^{14}\ \mathrm{Hz}}$. Cross-check: $c/\lambda_0 = (3.0\times10^{8})/(544\times10^{-9}) \approx 5.5\times10^{14}\ \mathrm{Hz}$, agreeing to rounding.

Step 3. Blue light, $480\ \mathrm{nm}$: its frequency is $f = c/\lambda = (3.0\times10^{8})/(480\times10^{-9}) = 6.25\times10^{14}\ \mathrm{Hz}$, which is above $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$ (equivalently $480\ \mathrm{nm} < 544\ \mathrm{nm}$). So blue light does eject electrons.

Step 4. Sodium yellow, $589\ \mathrm{nm}$: $f = (3.0\times10^{8})/(589\times10^{-9}) = 5.09\times10^{14}\ \mathrm{Hz}$, which is below $f_0$ (equivalently $589\ \mathrm{nm} > 544\ \mathrm{nm}$). So the sodium line cannot eject electrons from sodium — it is just past the threshold on the wrong side.

Competitive level — the intensity trap (JEE and NEET pattern)

Monochromatic light above threshold falls on a photocell and produces a saturation current of $8\ \mu\mathrm{A}$ and a stopping potential of $1.2\ \mathrm{V}$. Keeping the frequency fixed, the intensity is doubled. State the new saturation current and the new stopping potential, and find the maximum kinetic energy of the photoelectrons.

Step 1. Saturation current tracks intensity (Law 3): doubling the intensity at fixed frequency doubles the number of photoelectrons per second, so the saturation current becomes $2\times8 = \boxed{16\ \mu\mathrm{A}}$.

Step 2. Stopping potential depends only on frequency and work function (Law 4), never on intensity. It is therefore unchanged: $V_0 = \boxed{1.2\ \mathrm{V}}$. Every tempting answer that scales $V_0$ with brightness is wrong.

Step 3. Maximum kinetic energy from the stopping potential (equation 3): $KE_{\max} = eV_0 = 1.2\ \mathrm{eV} = 1.2\times1.60\times10^{-19} = \boxed{1.92\times10^{-19}\ \mathrm{J}}$. This too is unchanged by the brightening — the fastest electron carries the same energy before and after. Timing note: about 45 seconds. There is nothing to calculate in Steps 1 and 2 once you have internalised "intensity → number, frequency → energy"; only the last conversion $1.2\ \mathrm{eV} \to 1.92\times10^{-19}\ \mathrm{J}$ needs the calculator. Recognising that $V_0$ is fixed is worth full marks by inspection.

Competitive level — the classical wait (JEE Advanced pattern)

A caesium surface ($\varphi = 2.14\ \mathrm{eV}$) is lit by a source of intensity $1\ \mathrm{W\,m^{-2}}$. Modelling one atom as a disc of radius $1.0\times10^{-10}\ \mathrm{m}$, estimate the time a purely classical wave would need to feed one electron its escape energy, and compare with the observed emission delay.

Step 1. Collecting area of one atom: $A = \pi r^2 = \pi(1.0\times10^{-10})^2 = 3.14\times10^{-20}\ \mathrm{m^2}$. Power funnelled to it classically: $P = IA = (1)(3.14\times10^{-20}) = 3.14\times10^{-20}\ \mathrm{W}$.

Step 2. Escape energy in joules: $\varphi = 2.14\ \mathrm{eV} = 2.14\times1.60\times10^{-19} = 3.42\times10^{-19}\ \mathrm{J}$.

Step 3. Classical wait: $t = \varphi/P = (3.42\times10^{-19})/(3.14\times10^{-20}) = \boxed{10.9\ \mathrm{s}}$ — about eleven seconds. Experiment sees emission in under $10^{-9}\ \mathrm{s}$, so the classical estimate is too long by more than ten orders of magnitude — the decisive failure of the wave model of energy delivery. Timing note: about 90 seconds. The whole item is one area, one product, one division; the marks are in stating the conclusion — that the observed nanosecond delay is impossibly shorter than the classical seconds — not in the arithmetic.

Intensity versus frequency — the master trap. Brightness sets the saturation current (the number of electrons); frequency sets the maximum kinetic energy and the stopping potential (the speed of the fastest electron). Any question that changes the brightness and asks about the stopping potential is testing exactly this: $V_0$ does not move. On CBSE, ICSE, JEE, NEET, AP Physics 2, IB HL and A-Level 9702 this appears every year.

"Bright enough will always work." No. Below the threshold frequency $f_0$, no intensity and no exposure time ejects a single electron. If a problem gives light with $f < f_0$, the answer is zero photocurrent — full stop.

Threshold in wavelength has the inequality flipped. Higher frequency means shorter wavelength, so the emitting condition is $f > f_0$ but $\lambda < \lambda_0$. Students routinely write $\lambda > \lambda_0$ for emission and lose the mark. Always sanity-check: bluer light (shorter $\lambda$) is the one that ejects electrons.

Stopping potential is an energy, read through $e$. $eV_0 = KE_{\max}$, so a stopping potential of $1.5\ \mathrm{V}$ means $KE_{\max} = 1.5\ \mathrm{eV}$, not $1.5\ \mathrm{J}$. Convert with $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$ only when the answer is demanded in joules.

Do not quote the Einstein equation here as if it were derived. This article establishes the four laws and the classical failure; the equation $KE_{\max} = hf - \varphi$ and the slope $h/e$ belong to the next articles. If a proof of the $V_0$–$f$ line is asked, it needs the photon, which is not yet in hand.

Board scope. The photoelectric effect and its laws are on CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2 (Unit 15), IB HL (Theme E), and A-Level (CIE 9702). It is IB HL-only — Standard Level does not carry it — and IGCSE 0625 has no photoelectric content at all, so an IGCSE student should not be studying this page for their exam.

Texts. NCERT Physics Class 12, Chapter 11 ("Dual Nature of Radiation and Matter"), the sections on the photoelectric effect and its experimental laws · H. C. Verma, Concepts of Physics Vol. 2, the photoelectric-effect chapter · Halliday, Resnick and Walker, "Photons and Matter Waves".

Provenance. Representative constants. Recompute: caesium $\lambda_0 = 1240/2.14 = 579\ \mathrm{nm}$, $f_0 = (2.14\times1.60\times10^{-19})/(6.63\times10^{-34}) = 5.16\times10^{14}\ \mathrm{Hz}$; sodium $\lambda_0 = 1240/2.28 = 544\ \mathrm{nm}$, $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$; tungsten $\lambda_0 = 1240/4.5 = 276\ \mathrm{nm}$; zinc $\lambda_0 = 1240/4.3 = 288\ \mathrm{nm}$; $f(480\ \mathrm{nm}) = 6.25\times10^{14}\ \mathrm{Hz}$, $f(589\ \mathrm{nm}) = 5.09\times10^{14}\ \mathrm{Hz}$; classical time lag $t = 3.42\times10^{-19}/3.14\times10^{-20} = 10.9\ \mathrm{s}$ (and $\approx 126\ \mathrm{days}$ at $10^{-6}\ \mathrm{W\,m^{-2}}$); $KE_{\max} = eV_0 = 1.2\ \mathrm{eV} = 1.92\times10^{-19}\ \mathrm{J}$; slope $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$.

Prerequisites. Electron Emission & Work Function — where the free-electron sea, the surface barrier, and the work function $\varphi$ are set up · and The Nature of Electromagnetic Waves, for what "frequency" and "intensity" of light mean. Leads to. Planck's Quantum & the Photon, the next article, which introduces the light quantum $E = hf$ that resolves every failure catalogued here · and, for enrichment, the university seam The Photoelectric Effect and the Photon (go-deeper only, never an exam dependency here).