Einstein's Photoelectric Equation

Einstein's ledger runs quietly inside a startling amount of modern hardware. Every photomultiplier tube and image-intensifier — the night-vision goggle, the particle detector, the medical PET scanner — begins with a photon knocking an electron off a photocathode chosen for its low work function, exactly the caesium-and-antimony surfaces this article uses for numbers. The solar cell is a semiconductor cousin: a photon must carry at least the band-gap energy (the material's threshold) to free a charge carrier, which is why silicon panels ignore far-infrared light and why engineers stack materials of different gaps to catch more of the spectrum. Digital camera sensors count photoelectrons pixel by pixel; the "ISO" you set is really a gain on that photoelectron count. And the equation's deepest legacy is conceptual: it was the first hard evidence that light is granular, forcing the wave–particle duality that the very next module extends from radiation to matter. The same pair of fundamental constants, combined as $2e/h$ through the Josephson effect, underlies the modern voltage standard, and the same $hf - \varphi$ accounting, run in reverse and refined, governs the photoelectron spectroscopy (XPS, UPS) that maps the electronic structure of every new material in a chemistry lab.

Forget waves for a moment and picture light as a stream of tiny, indivisible parcels of energy. Each parcel — a photon — carries exactly $E = hf$, set only by the light's frequency $f$ and Planck's constant $h$. Brightness is not the energy per photon; brightness is the number of photons per second. A dazzling red beam is a torrent of low-energy photons; a dim violet beam is a trickle of high-energy ones. This single reframing is the hinge on which the whole effect turns, because it separates two quantities that a wave hopelessly tangles together: how energetic each hit is (frequency) versus how many hits arrive (intensity). Now watch one electron sitting near the surface of a metal. It is held inside by a small energy wall — the atoms' collective pull — and the height of that wall is the work function $\varphi$, the least energy the electron needs to break free. When a photon strikes, the electron does not sip a little energy from a passing wave and wait for more; it swallows the entire photon in one gulp or nothing at all. If that gulp $hf$ is smaller than the wall $\varphi$, the electron simply cannot get out — it jiggles and settles back, no matter how many such photons rain down. If $hf$ exceeds $\varphi$, the electron escapes immediately, carrying away the leftover energy as motion. There is the threshold, and there is the instantaneity, already visible in the picture before we write a single equation. The energy bookkeeping is a one-line ledger, and it is worth seeing as a physical balance before it is an algebraic one: the energy delivered ($hf$) equals the energy spent escaping ($\varphi$) plus the energy kept as motion ($KE$). Because different electrons sit at different depths and lose different amounts on the way out, the ones that escape with the most kinetic energy are those that started right at the surface and paid only the minimum $\varphi$. Their kinetic energy is the maximum the experiment can produce, $KE_{\max}$, and it is this quantity — not some average — that the stopping-potential measurement isolates.

The physical ledger becomes Einstein's equation the moment we write it in symbols. A photon delivers $hf$; the electron pays the escape cost $\varphi$; the remainder is the greatest kinetic energy a freed electron can carry. Because the electrons that escape with the most energy are the ones that paid only the minimum toll $\varphi$, we label that remainder $KE_{\max}$. The balance reads:

where: $KE_{\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant, $f$ is the frequency of the incident light, and $\varphi$ is the work function of the metal (a fixed property of its surface). The physical reading is a strict conservation statement: energy in equals energy spent escaping plus energy kept as motion. Two features are already decisive. First, $KE_{\max}$ depends on the frequency $f$, not on how bright the light is — brightness only sets how many electrons come out, never how fast. Second, if $hf < \varphi$ the right-hand side would be negative, which is physically impossible, so the electron never leaves: that is the threshold, built into the algebra. Set $KE_{\max} = 0$ and you find the exact borderline. The lowest frequency that can free any electron at all satisfies $hf_0 = \varphi$, defining the threshold frequency $f_0 = \varphi/h$ (and the corresponding threshold wavelength $\lambda_0 = hc/\varphi = 1240/\varphi[\mathrm{eV}]$ nm). Substituting $\varphi = hf_0$ back into equation (1) rewrites Einstein's law in its most revealing form:

where: $f_0$ is the threshold frequency below which no emission occurs and $f - f_0$ is the "excess" frequency above threshold. The physical reading is that only the frequency above $f_0$ does useful work: light exactly at threshold ($f = f_0$) liberates electrons with zero speed, and every hertz beyond that adds kinetic energy at the fixed rate $h$. For caesium ($\varphi = 2.14\ \mathrm{eV}$) the threshold sits at $\lambda_0 = 1240/2.14 = 579\ \mathrm{nm}$ (yellow-orange) and $f_0 = \varphi/h = 5.16\times10^{14}\ \mathrm{Hz}$; for sodium ($\varphi = 2.28\ \mathrm{eV}$), $\lambda_0 = 544\ \mathrm{nm}$ and $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$; for tungsten ($\varphi = 4.5\ \mathrm{eV}$), $\lambda_0 = 276\ \mathrm{nm}$ — deep in the ultraviolet, which is why tungsten needs UV where caesium responds to visible light.

How do you measure the kinetic energy of an electron you cannot see? You make it climb an electrical hill. Apply a voltage that pushes the ejected electrons back toward the metal, and make it steadily more negative until even the fastest electron is turned around just short of the collector — at that voltage the photocurrent falls to zero. This "just enough to stop them" voltage is the stopping potential $V_0$. An electron of charge $e$ moving against a potential difference $V_0$ loses energy $eV_0$, so stopping the fastest electron means its entire kinetic energy has been spent climbing the hill:

where: $e = 1.60\times10^{-19}\ \mathrm{C}$ is the electron's charge and $V_0$ is the stopping potential (in volts). The physical reading is a clean unit-conversion between an energy you cannot see and a voltage you can read off a meter. A neat consequence of measuring energies in electronvolts: since $1\ \mathrm{eV}$ is the energy $e\times 1\ \mathrm{V}$, the number of volts in $V_0$ equals the number of electronvolts in $KE_{\max}$. If $KE_{\max} = 0.96\ \mathrm{eV}$ then $V_0 = 0.96\ \mathrm{V}$ exactly — the meter reads the kinetic energy directly. Now combine the stopping-potential relation (3) with Einstein's equation (1). Substitute $KE_{\max} = eV_0$ into $KE_{\max} = hf - \varphi$ and divide through by $e$:

where: the coefficient $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$ multiplies the frequency and $\varphi/e$ is the work function expressed in volts. This is the equation of a straight line $y = mx + c$ with $y = V_0$ and $x = f$. Its slope is $h/e$, a combination of two fundamental constants that does not depend on which metal you use — every metal gives a line of the same tilt. The line crosses the frequency axis ($V_0 = 0$) at $f = f_0 = \varphi/h$, and if you extend it back to $f = 0$ it meets the voltage axis at $-\varphi/e$. So a single graph reveals two things at a glance: the slope hands you $h$ (the same for all metals), and the intercepts hand you the work function of the particular metal. This is the graph that Millikan measured, and the one exam boards from CBSE to A-Level love to test.

Einstein's one equation dissolves, one by one, the four observations that had made the photoelectric effect a scandal for wave theory. The first is the threshold. Classically, a wave of any frequency should eventually free an electron if you pour in enough energy; experimentally, light below $f_0$ frees nothing, ever. Equation (2) explains it exactly: a single photon must clear the wall by itself, and if $hf < \varphi$ no individual photon can, so pouring in more photons only means more failed attempts. The second is instantaneity. A wave would have to trickle energy into an electron over a measurable time; experiment finds emission within $10^{-9}\ \mathrm{s}$ even in near-darkness. The photon picture removes the wait entirely — the transaction is a single absorption event, complete the moment a photon arrives. The third riddle is that the electrons' energy depends on colour, not brightness. In equation (1), $KE_{\max} = hf - \varphi$ contains $f$ but not intensity; doubling the brightness doubles the number of photons and hence the number of electrons, but leaves each photon's energy — and so each electron's maximum kinetic energy — untouched. That is precisely what Lenard had found and what no wave model could stomach. The fourth is the mirror image: the saturation current is proportional to intensity. More photons per second (above threshold) means more electrons freed per second, so the current climbs with brightness even though the stopping potential does not budge. Four "impossibilities" become four corollaries of a single accounting rule — the mark of a genuinely deep idea rather than a fudge.

Board level — caesium under violet light (CBSE and ICSE pattern)

Caesium has a work function $\varphi = 2.14\ \mathrm{eV}$. Light of wavelength $400\ \mathrm{nm}$ falls on a clean caesium surface. Find (a) the energy of each photon in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping potential, and (d) the threshold wavelength of caesium. Use $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$.

Step 1. Photon energy: $E = 1240/\lambda = 1240/400 = \boxed{3.10\ \mathrm{eV}}$. Each violet photon carries $3.10\ \mathrm{eV}$, comfortably above caesium's $2.14\ \mathrm{eV}$ barrier, so emission will occur.

Step 2. Maximum kinetic energy from Einstein's equation: $KE_{\max} = hf - \varphi = 3.10 - 2.14 = \boxed{0.96\ \mathrm{eV}}$. This is the energy of the fastest electrons — the ones that started at the surface and paid only $\varphi$.

Step 3. Stopping potential: $eV_0 = KE_{\max}$, and since $KE_{\max}$ is $0.96\ \mathrm{eV}$, the number of volts equals the number of electronvolts, so $V_0 = \boxed{0.96\ \mathrm{V}}$. A retarding voltage of $0.96\ \mathrm{V}$ just stops the photocurrent.

Step 4. Threshold wavelength: at threshold the whole photon pays the barrier, $hc/\lambda_0 = \varphi$, so $\lambda_0 = 1240/\varphi = 1240/2.14 = \boxed{579\ \mathrm{nm}}$. Any light with wavelength longer than $579\ \mathrm{nm}$ (redder than yellow-orange) cannot free an electron from caesium, however intense. Cross-check: $f_0 = \varphi/h = (2.14 \times 1.60\times10^{-19})/(6.63\times10^{-34}) = 5.16\times10^{14}\ \mathrm{Hz}$, and $c/f_0 = (3.0\times10^8)/(5.16\times10^{14}) = 581\ \mathrm{nm}$, consistent with $579\ \mathrm{nm}$ to rounding.

Competitive level — recovering Planck's constant from two colours (JEE and NEET pattern)

A caesium photocell ($\varphi = 2.14\ \mathrm{eV}$) is illuminated first with $400\ \mathrm{nm}$ light and then with $500\ \mathrm{nm}$ light. Measure the two stopping potentials, use them to find the slope of the $V_0$–$f$ line, and hence estimate Planck's constant. Take $c = 3.0\times10^8\ \mathrm{m\,s^{-1}}$ and $e = 1.60\times10^{-19}\ \mathrm{C}$.

Step 1. Convert wavelengths to frequencies: $f_1 = c/\lambda_1 = (3.0\times10^8)/(400\times10^{-9}) = 7.5\times10^{14}\ \mathrm{Hz}$ and $f_2 = c/\lambda_2 = (3.0\times10^8)/(500\times10^{-9}) = 6.0\times10^{14}\ \mathrm{Hz}$.

Step 2. Photon energies and stopping potentials. At $400\ \mathrm{nm}$: $E_1 = 1240/400 = 3.10\ \mathrm{eV}$, so $V_{0,1} = E_1 - \varphi = 3.10 - 2.14 = 0.96\ \mathrm{V}$. At $500\ \mathrm{nm}$: $E_2 = 1240/500 = 2.48\ \mathrm{eV}$, so $V_{0,2} = 2.48 - 2.14 = 0.34\ \mathrm{V}$. Both photons clear the barrier, so both give real (positive) stopping potentials.

Step 3. Slope of the $V_0$–$f$ line from the two points: $\dfrac{V_{0,1} - V_{0,2}}{f_1 - f_2} = \dfrac{0.96 - 0.34}{(7.5 - 6.0)\times10^{14}} = \dfrac{0.62}{1.5\times10^{14}} = \boxed{4.13\times10^{-15}\ \mathrm{V\,s}}$. This should equal $h/e$; the accepted value is $4.14\times10^{-15}\ \mathrm{V\,s}$, and the tiny gap is the rounding baked into the $1240$ shortcut.

Step 4. Planck's constant is the slope times $e$: $h = (h/e)\times e = (4.13\times10^{-15})(1.60\times10^{-19}) = 6.6\times10^{-34}\ \mathrm{J\,s}$, recovering the tabulated $6.63\times10^{-34}\ \mathrm{J\,s}$. This is exactly Millikan's strategy — read $h$ off the slope of a line he expected to be curved. Timing note: about 90 seconds under exam pressure. Do not compute $\varphi$ twice; it cancels in the slope, which is the whole reason the two-point method is fast. Convert to frequencies once, subtract, divide — and remember the slope alone gives $h/e$, no work function needed.

Confusing what intensity and frequency each control. Frequency sets the energy of each photon (hence $KE_{\max}$ and $V_0$); intensity sets the number of photons per second (hence the saturation current). Almost every trick question swaps these. "Doubling the intensity doubles the stopping potential" is false; "doubling the intensity doubles the current" is true (above threshold).

Dropping the work function. Writing $KE_{\max} = hf$ instead of $hf - \varphi$ is the classic sign error's cousin. The photon's energy is not all available as kinetic energy — the electron must first pay $\varphi$ to escape. Only far above threshold is $KE_{\max} \approx hf$ a fair approximation.

Threshold in the wrong direction. Emission needs $f \ge f_0$, equivalently $\lambda \le \lambda_0$. Because wavelength runs opposite to frequency, the threshold wavelength is a maximum: light must be shorter than $\lambda_0$. Students routinely invert this and reject blue light while accepting red.

Units in the slope. The slope $h/e$ is $4.14\times10^{-15}\ \mathrm{V\,s}$, not $4.14\times10^{-15}\ \mathrm{J\,s}$ (that is $h$ itself in different units, and even then $h = 6.63\times10^{-34}\ \mathrm{J\,s}$). If your graph's slope comes out near $6.6\times10^{-34}$ you plotted $KE_{\max}$ (in joules) against $f$, whose slope is $h$; the $V_0$-versus-$f$ slope is $h/e$.

Board scope. Einstein's equation and the stopping-potential graph are core on CBSE, ICSE, JEE (Main and Advanced), NEET, AP Physics 2 (Unit 15), and A-Level (CIE 9702), and are carried by IB Physics at Higher Level (Theme E) — but IB Standard Level does not require the photoelectric equation, and IGCSE 0625 has no photoelectric content at all. Do not tell an IGCSE or IB-SL student to memorise this line.

Texts. NCERT Physics Class 12, Chapter 11 ("Dual Nature of Radiation and Matter"), the section on Einstein's photoelectric equation and the study of photoelectric effect · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave–Particle Duality" · D. C. Pandey / Halliday–Resnick–Walker treatments of the stopping-potential experiment and the $V_0$–$f$ graph.

Provenance. Standard results with locked anchor numbers. Recompute: $E(400\ \mathrm{nm}) = 1240/400 = 3.10\ \mathrm{eV}$; $KE_{\max} = 3.10 - 2.14 = 0.96\ \mathrm{eV}$; $V_0 = 0.96\ \mathrm{V}$; caesium $\lambda_0 = 1240/2.14 = 579\ \mathrm{nm}$ and $f_0 = \varphi/h = 5.16\times10^{14}\ \mathrm{Hz}$; sodium $\lambda_0 = 544\ \mathrm{nm}$, $f_0 = 5.50\times10^{14}\ \mathrm{Hz}$; tungsten $\lambda_0 = 276\ \mathrm{nm}$; slope $h/e = 4.14\times10^{-15}\ \mathrm{V\,s}$; two-point slope $(0.96-0.34)/(1.5\times10^{14}) = 4.13\times10^{-15}\ \mathrm{V\,s}$; $E(500\ \mathrm{nm}) = 2.48\ \mathrm{eV}$, $V_{0,2} = 0.34\ \mathrm{V}$; $E(656\ \mathrm{nm}) = 1.89\ \mathrm{eV}$ (below caesium threshold).

Prerequisites. The Photoelectric Effect & Its Laws — the four experimental facts (threshold, instantaneity, current $\propto$ intensity, $V_0 \propto$ frequency) that this equation must explain · Planck's Quantum & the Photon — where $E = hf$ is established. Leads to. Dual Nature of Radiation & Compton Scattering — the momentum side of the photon, and, for enrichment, the university seam on the quantum theory of the photoelectric effect and the photon (a go-deeper pointer, never an exam dependency here).