Einstein's photoelectric equation

In 1902 Philipp Lenard shone light on a metal plate and watched electrons fly off it, and every serious physicist of the day expected the escaping electrons to get more energetic as the light was made brighter. That is exactly what a wave should do — a bigger wave carries more energy, so a brighter wave should hit each electron harder. It did not happen. Turning up the brightness produced more electrons, not faster ones. Worse, below a certain colour of light, no electrons came out at all, no matter how long you waited or how intense the beam. Classical wave theory had no answer. Albert Einstein did, in 1905, in the same year he also explained Brownian motion and special relativity — and he did it with a single short equation that treats light not as a continuous wave but as a hail of discrete energy packets. That equation, $eV_0 = h\nu - \phi_0$ , is one of the shortest and most consequential lines in the entire Class 12 syllabus. It single-handedly rescues the photoelectric effect from paradox, it is directly testable on a metre-long graph you can sketch on graph paper, and it earned Einstein his only Nobel Prize — not for relativity, but for this. This article builds the equation from the photon idea, shows exactly why it fixes every experimental puzzle that broke the wave picture, works two board-style numbers all the way through, and turns the whole thing into an exam-ready toolkit: what to plug in, what units to use, and the traps that quietly cost marks every year.

Key takeaways

Light arrives in packets. Einstein proposed that light of frequency $\nu$ is made of photons, each carrying a fixed quantum of energy $E = h\nu$ , with $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ Planck's constant.
One photon, one electron. Photoemission is a one-to-one event: a single photon is absorbed by a single electron, never shared or accumulated over time.
The equation itself. $eV_0 = h\nu - \phi_0$ , equivalently $eV_0 = h(\nu-\nu_0)$ , links the stopping potential $V_0$ to the frequency $\nu$ , the work function $\phi_0 = h\nu_0$ , and the threshold frequency $\nu_0$ .
It explains the threshold. Below $\nu_0$ , a photon simply cannot supply $\phi_0$ , so no electron escapes — regardless of intensity or exposure time.
It explains intensity-independence. Brightness only changes how many photons arrive per second, not the energy of any one photon, so $KE_{max}$ never depends on intensity.
It is a straight line in disguise. Rearranged as $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$ , a plot of $V_0$ against $\nu$ is a straight line of slope $h/e$ — the experimental method used to measure $h$ independently of blackbody radiation.
Key idea: because each photon of light delivers a fixed quantum $h\nu$ to exactly one electron, the maximum photoelectron energy obeys $eV_0=h\nu-\phi_0$ , which fixes a threshold frequency, makes $KE_{max}$ independent of intensity, and turns a plot of $V_0$ against $\nu$ into a straight line whose slope is $h/e$ .

The core insight

Einstein's photoelectric equation $eV_0 = h\nu - \phi_0 = h(\nu - \nu_0)$ treats light as photons of energy $h\nu$ absorbed one at a time by one electron each, so the fastest photoelectron's kinetic energy rises linearly with frequency above a fixed threshold $\nu_0$ and is completely unaffected by how bright the light is.

Where it fits in the syllabus

"Einstein's photoelectric equation — particle nature of light" is named explicitly in the NCERT Class 12 unit "Dual Nature of Radiation and Matter," alongside the qualitative Hertz/Lenard history and the experimental trends of the photoelectric effect. This equation is the quantitative heart of that chapter: everything before it (threshold frequency, stopping potential, saturation current, the failure of wave theory) is observation, and everything the equation explains is the payoff. The 2025-26 CBSE syllabus keeps this topic firmly in scope and testable with numericals; it does not extend into Compton scattering, X-ray production, or a detailed derivation of the photoelectric current itself, which belong to higher-level or competitive-exam physics and are correctly left out here.

The physical picture

Einstein extended Max Planck's idea — that a hot object exchanges energy with radiation only in discrete lumps — to light itself. In this picture, light of frequency $\nu$ is not a continuous wave spreading its energy smoothly through space, but a stream of particle-like packets called photons, each one carrying a fixed quantum of energy:

E = h\nu

where: $E$ is the energy of a single photon (joules), $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ is Planck's constant, and $\nu$ is the frequency of the light (hertz). The physical reading is blunt: a beam of blue light and a beam of red light of the same power contain a different number of photons, because each blue photon is individually more energetic than each red one ( $E \propto \nu$ ). Brightness (intensity) is a headcount of photons arriving per second; colour (frequency) is the energy carried by each individual photon. These are two completely independent knobs, and confusing them is the single biggest reason classical wave theory failed here. When such a photon strikes a metal surface, photoemission proceeds as a strict one-to-one interaction: one photon is absorbed entirely by one electron, instantly, with no sharing among several electrons and no gradual build-up over multiple photon arrivals. Part of that absorbed energy is spent doing work to free the electron from the metal — overcoming the attractive pull of the positive ions at the surface, quantified by the work function $\phi_0$ , the minimum energy an electron needs to just escape. Whatever energy is left over becomes the electron's kinetic energy as it leaves the surface. Electrons deeper inside the metal lose some of this leftover energy to collisions on the way out, so real photoelectrons emerge with a whole spread of kinetic energies from zero up to a single maximum, achieved only by electrons that were sitting right at the surface and lost nothing extra:

KE_{max} = h\nu - \phi_0

where: $KE_{max}$ is the maximum kinetic energy any photoelectron can carry (joules or eV), $h\nu$ is the energy delivered by one incident photon, and $\phi_0$ is the work function of the particular metal (a fixed material property, typically a few eV). Experimentally, $KE_{max}$ is measured by applying a reverse (retarding) voltage across the photocell just large enough to stop even the fastest electrons from reaching the collector — the stopping potential $V_0$ . Since work done against this retarding field equals the kinetic energy it just barely overcomes, $eV_0 = KE_{max}$ , and substituting gives Einstein's photoelectric equation in its standard, examinable form:

eV_0 = h\nu - \phi_0

where: $e = 1.6\times10^{-19}\ \text{C}$ is the electron's charge, $V_0$ is the stopping potential (volts), $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function. Because no emission can occur once $h\nu$ drops below $\phi_0$ , there must exist a threshold frequency $\nu_0$ at which the photon energy exactly equals the work function, $\phi_0 = h\nu_0$ ; substituting this back gives the equally useful second form:

eV_0 = h(\nu - \nu_0)

Both forms say the identical thing: the maximum energy a photoelectron can carry grows linearly with how far the incident frequency sits above the metal's own threshold, and nothing else about the light matters.

Figure

A graph of stopping potential on the vertical axis against frequency of light on the horizontal axis, showing a straight rising line that crosses the frequency axis at the threshold frequency nu naught and has slope h over e, with the same line shifted right and parallel for a metal of larger work function

Einstein's photoelectric equation, derived

Derivation ABuilding Einstein's photoelectric equation from the photon idea

The goal of this derivation is to combine three separate, independently justified physical facts — the photon energy formula, energy conservation at the metal surface, and the operational meaning of the stopping potential — into the single testable relation $eV_0 = h\nu - \phi_0$ , without invoking anything beyond algebra. Each step is a statement most students already accept individually; the derivation is really about assembling them correctly and identifying exactly which electrons the final formula describes.

Step 1. Assign the energy carried by one photon. By Einstein's quantum hypothesis, a photon of light of frequency $\nu$ carries energy $E = h\nu$ , and — this is the crucial particle-like assumption — that entire packet is delivered to at most one electron in a single absorption event; it cannot be split between two electrons, nor can two photons combine their energy on the same electron under ordinary illumination.

Step 2. Apply conservation of energy to the electron that absorbs it. The absorbed energy $h\nu$ has exactly two places to go: some fraction is spent as work done against the forces holding the electron inside the metal, and the remainder appears as the electron's outgoing kinetic energy. For the least tightly bound electrons — those effectively already at the surface — the minimum work required is by definition the work function $\phi_0$ , so these electrons carry away the largest possible share of leftover energy: $KE_{max} = h\nu - \phi_0$ . Electrons drawn from deeper inside the metal do additional work escaping and so leave with less than $KE_{max}$ , which is why real photocurrents show a whole distribution of electron energies up to this one ceiling.

Step 3. Translate $KE_{max}$ into something a voltmeter can read. The stopping potential $V_0$ is defined as the smallest retarding voltage that reduces the photocurrent to exactly zero, i.e. the voltage that does just enough negative work $eV_0$ on the fastest electron to cancel its kinetic energy at the collector. Equating this retarding work to the kinetic energy it exactly cancels gives $eV_0 = KE_{max}$ .

Step 4. Substitute Step 2 into Step 3 to get the working equation. Combining $KE_{max} = h\nu - \phi_0$ with $eV_0 = KE_{max}$ gives $eV_0 = h\nu - \phi_0$ . Since no photoemission is physically possible once the right-hand side would turn negative, there is a threshold frequency $\nu_0$ defined by $h\nu_0 = \phi_0$ exactly, at the boundary where $KE_{max}$ first reaches zero; substituting $\phi_0 = h\nu_0$ recasts the same equation as $eV_0 = h(\nu-\nu_0)$ , an equally valid and often more convenient form when $\nu_0$ is known directly rather than $\phi_0$ . Both forms are algebraically identical statements of energy conservation for the single most energetic photoelectron produced at frequency $\nu$ ; every numerical problem in this topic is just this equation solved for whichever one quantity is unknown.

Worked examples

Board level - photon energy, KE, and stopping potential from a wavelength

Problem: The work function of a metal is 2.14 eV. Light of wavelength 400 nm falls on it. Find (a) the energy of each incident photon in eV, (b) the maximum kinetic energy of the emitted photoelectrons in eV, and (c) the stopping potential.

Step 1. Convert wavelength to photon energy using $E = h\nu = \dfrac{hc}{\lambda}$ , with $h = 6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ , $c = 3\times10^8\ \text{m/s}$ , and $\lambda = 400\times10^{-9}\ \text{m}$ :

E = \frac{(6.63\times10^{-34})(3\times10^8)}{400\times10^{-9}} = \frac{1.989\times10^{-25}}{4\times10^{-7}} = 4.97\times10^{-19}\ \text{J}

Step 2. Convert this photon energy into electron-volts, since the work function is already given in eV, so that every term in the next step is in the same unit. Using $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ :

E = \frac{4.97\times10^{-19}}{1.6\times10^{-19}} \approx 3.11\ \text{eV}

Step 3. Apply Einstein's equation directly in eV, since $\phi_0 = 2.14$ eV is already in the matching unit:

KE_{max} = E - \phi_0 = 3.11 - 2.14 = \mathbf{0.97\ eV}

Step 4. Read off the stopping potential. Because $eV_0 = KE_{max}$ and $KE_{max}$ is expressed in eV, the numerical value of $V_0$ in volts equals the numerical value of $KE_{max}$ in eV:

\mathbf{V_0 = 0.97\ V}

Competitive level - extracting h and the work function from two data points

Problem: For a given metal, the stopping potential is 0.6 V for light of frequency $6\times10^{14}$ Hz, and 1.4 V for light of frequency $7.5\times10^{14}$ Hz. Use these two data points to find (a) an experimental value of Planck's constant, and (b) the work function of the metal, and (c) the threshold frequency.

Step 1. Write Einstein's equation once for each data point, treating $h$ and $\phi_0$ as the two unknowns of a pair of simultaneous linear equations:

eV_{01} = h\nu_1 - \phi_0 \qquad eV_{02} = h\nu_2 - \phi_0

Step 2. Subtract the first equation from the second to eliminate the unknown $\phi_0$ in one move, exactly the trick used whenever a $V_0$ -vs- $\nu$ graph is turned into a slope:

e(V_{02}-V_{01}) = h(\nu_2-\nu_1)

h = \frac{e(V_{02}-V_{01})}{\nu_2-\nu_1} = \frac{(1.6\times10^{-19})(1.4-0.6)}{(7.5-6)\times10^{14}} = \frac{(1.6\times10^{-19})(0.8)}{1.5\times10^{14}}

\mathbf{h \approx 8.53\times10^{-34}\ J \cdot s}

This is somewhat higher than the accepted value $6.63\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ , which is expected here since the numbers were deliberately rounded for a clean classroom calculation — the method, not this particular decimal, is what an exam is testing.

Step 3. Feed this experimental $h$ back into either original equation to isolate the work function; using the first data point:

\phi_0 = h\nu_1 - eV_{01} = (8.53\times10^{-34})(6\times10^{14}) - (1.6\times10^{-19})(0.6)

\phi_0 = 5.12\times10^{-19} - 0.96\times10^{-19} = 4.16\times10^{-19}\ \text{J} \approx \mathbf{2.6\ eV}

Step 4. Find the threshold frequency from $\phi_0 = h\nu_0$ , i.e. $\nu_0 = \phi_0/h$ :

\nu_0 = \frac{4.16\times10^{-19}}{8.53\times10^{-34}} \approx \mathbf{4.88\times10^{14}\ Hz}

Notice this is comfortably below both given frequencies, exactly as it must be for emission to have occurred at either one. Timing note: this two-point elimination method (subtract to kill $\phi_0$ , solve for $h$ , back-substitute for $\phi_0$ , then $\nu_0 = \phi_0/h$ ) is a standard four-line pattern worth memorising cold — in a competitive exam it should take under ninety seconds from reading the numbers to a boxed answer, leaving time in hand for the harder conceptual distractors on the same paper.

Common traps

Writing $KE_{max} = h\nu - \phi_0$ and treating it as the kinetic energy of every photoelectron, rather than the ceiling value — actual emitted electrons range continuously from $0$ up to $KE_{max}$ .
Mixing units mid-calculation: combining a photon energy already converted to eV with a work function still sitting in joules (or the reverse) without converting both to the same unit before subtracting.
Forgetting that $\nu$ in $E = h\nu$ is a frequency, and plugging a wavelength directly into the formula instead of first converting through $\nu = c/\lambda$ .
Trying to apply the equation below threshold frequency, where $h\nu < \phi_0$ would force a negative $KE_{max}$ — physically that regime simply has zero emission, not a negative kinetic energy.
Assuming that increasing the intensity of light below the threshold frequency will eventually trigger emission — it will not, because intensity only changes photon count per second, never the energy of an individual photon.
On the two-point slope method, forgetting to convert the frequency difference into the same power-of-ten scale before dividing, which silently shifts the computed $h$ by orders of magnitude.

You should be able to…

State Einstein's photoelectric equation in both forms, $eV_0 = h\nu - \phi_0$ and $eV_0 = h(\nu-\nu_0)$ , and explain what every symbol in it physically means.
Explain, using the equation, why a threshold frequency must exist and why no amount of intensity can substitute for it.
Explain why $KE_{max}$ is completely independent of the intensity of the incident light, and why that fact alone refuted classical wave theory.
Solve for any single unknown quantity — photon energy, work function, threshold frequency, stopping potential, or $KE_{max}$ — given the rest, including converting between wavelength and frequency where needed.
Use two stopping-potential/frequency data points to extract an experimental value of $h$ and the work function, keeping every quantity in consistent units throughout.
Interpret a $V_0$ -vs- $\nu$ graph: identify the slope as $h/e$ , the frequency intercept as $\nu_0$ , and the voltage intercept as $-\phi_0/e$ .

Ask the tutor: Want guided practice converting between wavelength, frequency, and photon energy, or working backward from a $V_0$ -vs- $\nu$ graph to pin down $h$ and $\phi_0$ for an unfamiliar metal?

If a photon just gives its whole energy to one electron, why don't all the photoelectrons come out with the exact same kinetic energy?

Because $KE_{max} = h\nu - \phi_0$ describes only the best case — an electron sitting right at the metal's surface that loses nothing extra escaping. Most photoelectrons start out somewhere below the surface, and as they travel outward through the metal they lose some of their newly gained energy to collisions with the lattice of metal ions before finally breaking free. That's why real photocurrents show a whole spread of kinetic energies from zero up to $KE_{max}$ , not one single value. Only the fastest, surface-level electrons actually reach the theoretical maximum the equation predicts.

Why does a stronger beam of red light below threshold still refuse to eject any electrons, even after hours of exposure?

Because emission is a strictly one-photon-one-electron event, not a slow accumulation of energy the way heating water is. Each red photon individually carries only $h\nu$ , and if that is less than $\phi_0$ , no single electron can ever collect enough energy to escape — it cannot borrow a little energy from one photon now and a little more from the next photon later, because photons interact with electrons essentially instantaneously and independently. A million too-weak photons arriving one after another are still each individually too weak; time and total intensity are irrelevant once $\nu < \nu_0$ .

Is the work function the same number for every metal, and where does it actually come from physically?

No — the work function $\phi_0$ is a fixed property of each particular metal, typically a few electron-volts, and it varies from metal to metal because it reflects how tightly that metal's positive ion lattice holds on to its outermost electrons. Metals with loosely bound valence electrons (like sodium or caesium) have small work functions and low threshold frequencies, useful in devices needing photoemission from visible light, while metals that bind their electrons more tightly need higher-frequency, more energetic photons before any emission at all is possible.

On a V0-versus-frequency graph, what exactly do the slope and the two intercepts tell us, and why is that experiment such a big deal?

Rearranging Einstein's equation as $V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}$ shows it is literally the equation of a straight line: the slope equals $h/e$ , the frequency-axis intercept is the threshold frequency $\nu_0$ , and the voltage-axis intercept is $-\phi_0/e$ . Robert Millikan spent about a decade painstakingly measuring exactly this straight line for several metals and found the slope gave a value of $h$ matching Planck's constant almost perfectly — turning Einstein's 1905 hypothesis from an elegant guess into a directly, independently verified physical constant, and it is the reason Einstein (not Millikan) received the 1921 Nobel Prize specifically for this equation.