Photoelectric Effect

Beginner

Shine light on a clean metal plate inside a vacuum tube and, with the right light, electrons jump out of its surface; a meter in the circuit then shows a current. This is the photoelectric effect. Let us collect what experiments show, one observation at a time.

Observation 1: colour decides. In the animation a bright red lamp floods the sodium plate and nothing happens, however long we wait. Dim violet light works at once.
Observation 2: no waiting. Even very faint violet light frees electrons the instant it arrives.
Observation 3: brightness decides how many. Brighter violet light ejects more electrons, but none of them faster.
Observation 4: bluer means faster. Ultraviolet light makes electrons leap out with more speed than violet.

A light wave spreading smoothly over the plate explains only the third observation. Einstein's answer, in 1905: light arrives in tiny packets of energy called photons, and the colour sets the energy of each packet. Red photons carry little, violet ones more, ultraviolet ones more still. An electron takes one photon at a time, all or nothing. To get out it must pay a fixed “exit fee”. A red photon cannot pay it, and more red photons do not help, just as a crowd in which each person holds one coin cannot buy a five-coin ticket. Whatever is left after the fee becomes the electron's speed.

Photomultiplier tubes use this effect to count single photons, and camera sensors and solar panels rely on a closely related effect. The photoelectric effect itself earned Einstein his Nobel Prize.

Check yourself: ultraviolet light frees electrons from a metal, and you make it twice as bright. Do the electrons come out faster, or do more of them come out? Answer: more of them, with the same top speed.

Intermediate

What the experiment measures

Two metal electrodes sit in an evacuated glass tube: a plate that the light falls on, here sodium, and a collector joined to it through a battery and an ammeter. In the dark the meter reads zero, because nothing carries charge across the vacuum. Light of a suitable colour makes a current flow at once: electrons, called photoelectrons, leave the lit plate and reach the collector.

The experimenter varies two properties of the light: its frequency $f$, seen as colour, and its intensity, the power it delivers per square metre. Two things are measured: the photocurrent $I$, which counts the electrons arriving each second, and the energy of the fastest electrons. Contrary to the wave picture, whether electrons come out, and how fast the fastest ones move, depends only on the frequency; the intensity sets nothing but the current.

Light in packets: the photon

In 1905 Einstein proposed that light is absorbed, and travels, in packets of energy now called photons. A photon of frequency $f$, or wavelength $\lambda =c/f$, carries the energy
$E=hf=\frac{hc}{\lambda }$
where $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is Planck's constant and $c$ the speed of light. Brighter light of one colour means more photons per second, not bigger ones.

Atomic physics measures such energies in electronvolts, the energy an electron gains across 1 V: $1\mathrm{eV}=1.602\times 10^{-19}\mathrm{J}$. With the product
$hc\approx 1240\mathrm{eV}\cdot \mathrm{nm}$
a photon's energy in electronvolts is 1240 divided by its wavelength in nanometres: 3.10 eV for violet light at 400 nm.

The energy balance: Einstein's equation

Conduction electrons move freely inside a metal, but freeing even the most loosely held one costs a definite energy. That energy is the work function $\phi$: about 2.1 eV for caesium, 2.28 eV for sodium, 4.3 eV for zinc and 5.1 eV for gold. An electron absorbs a whole photon or nothing. If $hf$ exceeds $\phi$, the electron pays its way out and keeps the rest as kinetic energy:
$K{E}_{\max }=hf-\phi$

The word “maximum” matters: only an electron that starts from the highest occupied energy and loses nothing on its way out leaves with all of $hf-\phi$. Electrons from lower levels, or from deeper inside, emerge with less, anything from zero up to $K{E}_{\max }$.

If $hf$ is smaller than $\phi$, no single photon can free an electron, and an electron cannot save up several photons: it passes its small gain to the metal as heat long before another photon arrives. The boundary case $hf=\phi$ defines the threshold frequency and the threshold wavelength:

• $f_0=\frac{\phi }{h}$
• ${\lambda }_0=\frac{hc}{\phi }$

The threshold is a minimum frequency but a maximum wavelength: light of longer wavelength than ${\lambda }_0$ frees nothing, however intense it is.

Worked example: the sodium plate in the animation

The plate in the animation is sodium, $\phi =2.28\mathrm{eV}$, so ${\lambda }_0=1240/2.28\approx 544\mathrm{nm}$, in the green, and $f_0\approx 5.51\times 10^{14}\mathrm{Hz}$. The bright red lamp at 700 nm gives $hf=1240/700=1.77\mathrm{eV}$ and $hf-\phi =1.77-2.28=-0.51\mathrm{eV}$. None of its 14 photons frees an electron, and a red tag reads no emission.

Violet light at 400 nm gives $hf=3.10\mathrm{eV}$ and $K{E}_{\max }=3.10-2.28=0.82\mathrm{eV}$; ultraviolet at 250 nm gives $hf=4.96\mathrm{eV}$ and $K{E}_{\max }=2.68\mathrm{eV}$. These are the values on the $\lambda$, $hf$ and $hf-\phi$ readouts. Since $0.82\mathrm{eV}=1.31\times 10^{-19}\mathrm{J}$, the fastest violet electrons move at $v=\sqrt{2K{E}_{\max }/m}\approx 5.4\times 10^{5}m/s$; the fastest ultraviolet ones reach $9.7\times 10^{5}m/s$.

The blue velocity arrows show this. One electron in three leaves with $K{E}_{\max }$; the others lost energy inside the metal and keep 45 % or 75 % of it, so their arrows are shorter. The fastest ultraviolet arrow is $\sqrt{2.68/0.82}\approx 1.8$ times as long as the fastest violet one. When the violet light turns bright, nine photons free nine electrons instead of three, yet the readout stays at 0.82 eV.

Worked example: a zinc plate under ultraviolet

A clean zinc plate, $\phi =4.3\mathrm{eV}$, is lit by the ultraviolet line of a mercury lamp at 254 nm. Does visible light work on it, how fast are the fastest electrons, and how large is the current?

Zinc's threshold is ${\lambda }_0=1240/4.3\approx 288\mathrm{nm}$, already in the ultraviolet, so the violet light that works on sodium frees nothing here. At 254 nm, $hf=1240/254=4.88\mathrm{eV}$ and $K{E}_{\max }=4.88-4.3=0.58\mathrm{eV}$: the fastest electrons move at about $4.5\times 10^{5}m/s$.

Suppose 1.0 mW of this light reaches the plate. At $4.88\mathrm{eV}=7.82\times 10^{-19}\mathrm{J}$ per photon, about $1.3\times 10^{15}$ photons arrive each second. A bare metal converts them poorly: if one in a thousand frees an electron, $1.3\times 10^{12}$ electrons leave each second and the current is $I\approx 1.3\times 10^{12}\cdot 1.602\times 10^{-19}\approx 2\times 10^{-7}\mathrm{A}$, or 0.2 µA. Doubling the lamp's power doubles the current but leaves the 0.58 eV unchanged.

Reading the energy diagram and the graphs

An energy diagram draws an electron's energy as height. The electrons fill the metal up to a top level, and an electron at rest just outside lies $\phi$ higher. A photon lifts one electron by exactly $hf$. For violet light, the 3.10 eV splits into 2.28 eV to climb out and 0.82 eV of motion; the red 1.77 eV falls 0.51 eV short, and the electron's gain ends up as heat.

For one colour above the threshold, the photocurrent against intensity is a straight line through the origin, while $K{E}_{\max }$ against intensity is flat. $K{E}_{\max }$ against frequency is zero up to $f_0$ and then a straight line of slope $h$: the same slope for every metal, starting from each metal's own $f_0=\phi /h$. The advanced view of the animation draws this graph.

Common mistakes

• Brighter light does not make faster electrons: tripling the violet light in the animation frees three times as many electrons, each still with at most 0.82 eV.
• Waiting does not help below the threshold: the red lamp could shine on sodium for hours, and each 1.77 eV photon would still fall short of 2.28 eV.
• The work function is subtracted, not added: violet photons on sodium give $3.10-2.28=0.82\mathrm{eV}$, not 5.38 eV.
• The threshold belongs to the metal, not to the light: 400 nm light frees electrons from sodium but not from zinc, whose threshold is 288 nm.
• Convert electronvolts into joules before using $KE=\frac{1}{2}m{v}^2$: 0.82 eV is $1.31\times 10^{-19}\mathrm{J}$ and gives 540 km/s, while putting in 0.82 directly gives $1.3\times 10^{15}m/s$, faster than light.

Where you meet it

Photomultiplier tubes count single photons in medical scanners, particle detectors and telescopes, multiplying each photoelectron about a million times. Night-vision intensifiers work the same way, and spacecraft in sunlight charge to a few volts positive as ultraviolet light frees electrons from their skin. Solar cells and camera sensors use the internal photoelectric effect, in which the photon lifts an electron inside a semiconductor rather than out of it.

Quanta of $hf$ run through the rest of modern physics. In Atomic Models: the Bohr Atom, atoms emit and absorb photons that match the gap between two energy levels; Semiconductors, Diodes and LEDs turns the effect around; Wave–Particle Duality gives electrons a wavelength too.

Advanced

Measuring the energy: the stopping potential

The photoelectric effect is settled by two measurements: how many electrons leave a lit metal each second, and how much kinetic energy the fastest of them carry. The first is simply a current. The second needs a trick, the stopping potential, and Figure 1 shows how it works.

Legend

$I$	Photocurrent: the charge per second carried by the electrons that reach the collector.	$\mathrm{A}$
$V$	Voltage of the collector relative to the lit plate; negative values push the electrons back.	$\mathrm{V}$
$V_0$	Stopping potential for the violet light: the current vanishes at $V=-V_0$.	$\mathrm{V}$
$V_0^{\prime }$	Stopping potential for the ultraviolet light; larger than $V_0$ because each ultraviolet photon carries more energy.	$\mathrm{V}$
$I_1$	Photocurrent in dim violet light; its plateau is the saturation current.	$\mathrm{A}$
$I_2$	Photocurrent in bright violet light, with three times as many photons per second as for $I_1$.	$\mathrm{A}$
$I_3$	Photocurrent in ultraviolet light, drawn with the most photons per second.	$\mathrm{A}$

The apparatus is a vacuum tube with two electrodes: a sodium plate that the light falls on, and a collector facing it. A voltage $V$ is applied to the collector, measured relative to the lit plate, and a meter records the photocurrent $I$, the charge per second carried by the electrons that reach the collector. Figure 1 shows three runs: dim violet light ($I_1$), brighter light of the same violet colour ($I_2$) and ultraviolet light ($I_3$).

Start on the right of Figure 1, where $V$ is positive. The collector attracts the electrons, every electron that leaves the plate is collected, and each curve levels off at a plateau, the saturation current. A plateau counts electrons per second, so for light of one colour it is proportional to the number of photons arriving per second: $I_2$, with three times as many photons as $I_1$, has a plateau three times as high. The ultraviolet run is drawn with the most photons per second, so its plateau is the highest. A plateau tells you how many electrons leave, never how fast they go.

Now move left, where $V$ is negative. The collector now repels the electrons, and to arrive an electron must climb a potential-energy hill of height $e|V|$, where $e=1.602\times 10^{-19}\mathrm{C}$ is the elementary charge. Only an electron that leaves the plate with more kinetic energy than this gets over. The electrons leave with a spread of energies and directions (Figure 2 shows where the spread comes from), so as $V$ is made more negative the slowest are turned back first and the current falls gradually. It reaches zero at $V=-V_0$, where even the fastest electrons, those with the largest kinetic energy $K{E}_{\max }$, are just turned back:
$e{V}_0=K{E}_{\max }$

The voltage $V_0$ is the stopping potential. It turns an energy measurement into a voltage reading: an electron stopped by exactly 1 V had 1 eV of kinetic energy, where the electronvolt, the energy an electron gains across 1 V, is $1.602\times 10^{-19}\mathrm{J}$. For sodium, $V_0=0.82\mathrm{V}$ under violet light of 400 nm and $2.68\mathrm{V}$ under ultraviolet light of 250 nm, the values on the animation's $V_0$ readout. The fastest electrons therefore carry 0.82 eV and 2.68 eV.

Now compare where the curves in Figure 1 reach zero. The two violet curves differ in height by a factor of three, yet both reach zero at the same $-V_0$. The ultraviolet curve $I_3$ reaches zero at $-V_0^{\prime }$, further to the left, because its electrons are faster. The same holds for any set of runs: every intensity of one colour stops at the same voltage, and each colour at its own. Intensity sets the current; frequency sets the energy. That is the whole puzzle in one figure, and the rest of this explanation shows why.

Energy conservation for one photon

Why should the energy depend on the colour alone? Einstein's answer, in 1905, was that light is absorbed in whole quanta, photons, each carrying the energy $hf$, where $f$ is the frequency of the light and $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is Planck's constant. Figure 2 follows the energy of one photon from its absorption until its electron is outside the metal.

Legend

$E$	Energy of an electron.	$\mathrm{J}$
$E_F$	Fermi level: the highest occupied electron state in the metal.	$\mathrm{J}$
$E_0$	Vacuum level: the energy of an electron at rest just outside the metal.	$\mathrm{J}$
$\phi$	Work function: the gap from $E_F$ up to $E_0$, the least energy that frees an electron.	$\mathrm{J}$
$hf$	Energy of the absorbed photon, the same for both electrons.	$\mathrm{J}$
$\epsilon$	How far below the Fermi level the second electron starts: the extra energy it must spend to get out.	$\mathrm{J}$
$K{E}_{\max }$	Kinetic energy, once outside, of the electron that starts on the Fermi level: $hf-\phi$.	$\mathrm{J}$
$KE$	Kinetic energy, once outside, of the electron that starts $\epsilon$ lower: $hf-\phi -\epsilon$.	$\mathrm{J}$

The vertical axis of Figure 2 is the energy $E$ of an electron. Inside a metal the conduction electrons fill the available states from the bottom up to a highest occupied level, the Fermi level $E_F$; the shaded band is the filled states. An electron at rest just outside the surface has the energy of the vacuum level $E_0$. The gap between the two, $\phi =E_0-E_F$, is the work function: the least energy that takes an electron out of the metal. For sodium it is 2.28 eV.

Absorbing a photon raises an electron by exactly $hf$: the purple arrows, equal for both electrons. Follow the electron on the left first. It starts on the Fermi level, so it needs only $\phi$ to reach the vacuum level. If it loses nothing on the way out, it keeps the remainder as kinetic energy, the blue arrow above $E_0$. No electron can do better, so this is the maximum:
$K{E}_{\max }=hf-\phi$

Now follow the electron on the right. It starts an energy $\epsilon$ below the Fermi level, so it must spend $\epsilon$ more to get out, and the same photon leaves it with only
$KE=hf-\phi -\epsilon$
Its blue arrow is shorter by $\epsilon$. Electrons from deeper states, and electrons that lose energy in collisions before they reach the surface, come out slower still: that is the spread of energies that made the curves of Figure 1 fall gradually. Setting $\epsilon =0$ gives back $K{E}_{\max }=hf-\phi$. Einstein's equation is energy conservation for one photon and one electron, nothing more; the new physics is entirely in the claim that light arrives in quanta of size $hf$.

The arrows are drawn to scale for ultraviolet light on sodium: $hf=4.96\mathrm{eV}$ and $\phi =2.28\mathrm{eV}$ give $K{E}_{\max }=2.68\mathrm{eV}$, exactly the $e{V}_0$ measured in Figure 1, and the electron that starts $\epsilon =0.9\mathrm{eV}$ lower leaves with $1.78\mathrm{eV}$.

One detail of this energy balance is easy to miss: a free electron cannot absorb a photon. To see why, let an electron of mass $m$ start at rest and take in a photon of energy $E$, which also carries momentum $E/c$. The electron would then have energy $m{c}^2+E$ and momentum $p=E/c$. But relativity fixes a particle's energy by its momentum: the square of the energy must be $(m{c}^2{)}^2+(pc{)}^2=(m{c}^2{)}^2+E^2$. The square of $m{c}^2+E$ is larger than that by $2m{c}^{2}E$, so energy and momentum cannot both be conserved. Something else must take up the momentum, the crystal lattice or the surface, so the effect belongs to electrons bound in a metal, not to isolated electrons.

Nor is the work function a constant special to light. The same barrier controls thermionic emission, in which heat frees the electrons: random thermal motion gives a few of them enough energy to climb over it, and the current is governed by the factor $e^{-\phi /kT}$, where $k$ is Boltzmann's constant and $T$ the absolute temperature. At room temperature $kT\approx 0.025\mathrm{eV}$, and for sodium the factor is below $10^{-38}$: no emission at all. A tungsten filament ($\phi \approx 4.5\mathrm{eV}$) at 2500 K has $kT\approx 0.22\mathrm{eV}$, and the factor reaches about $10^{-9}$, enough for a useful current. Heat pays the exit fee by chance, a few electrons at a time; a photon pays it in one go.

The straight line and its slope

Einstein's equation makes a sharp, testable prediction: plotted against frequency, $K{E}_{\max }$ is a straight line whose slope is the same for every metal. Figure 3 draws that prediction, which is also how $h$ was first measured with light.

Legend

$K{E}_{\max }$	Largest kinetic energy of the photoelectrons, $hf-\phi$.	$\mathrm{J}$
$f$	Frequency of the light.	$\mathrm{Hz}$
$f_0$	Threshold frequency of sodium, $\phi /h$: where its line meets the $f$ axis.	$\mathrm{Hz}$
$f_0^{\prime }$	Threshold frequency of zinc, larger because its work function is larger.	$\mathrm{Hz}$
$\phi$	Work function of sodium: the dashed continuation cuts the energy axis at $-\phi$.	$\mathrm{J}$
$\Delta f$	A step along the frequency axis, taken anywhere on the line.	$\mathrm{Hz}$
$h\Delta f$	Rise of the line over that step, the same everywhere: the slope of the line is $h$.	$\mathrm{J}$

Figure 3 plots $K{E}_{\max }=hf-\phi$ against the frequency $f$ of the light; the thick line is sodium. Look first at the small triangle on it: a step $\Delta f$ to the right raises the line by $h\Delta f$, and it does so wherever on the line the step is taken. That is what a slope of $h$ means.

Next, look where the line meets the axes. It cuts the frequency axis at the threshold frequency $f_0=\phi /h$, where a photon has just enough energy to free an electron, with nothing to spare. Below $f_0$ no electrons are emitted, however bright the light, so there is nothing to measure. Continued below $f_0$ (dashed), the line is an extrapolation, not a measurement; it cuts the energy axis at $-\phi$, a depth $\phi$ below the frequency axis, and reading off that intercept is one way to find the work function.

The thin line belongs to a metal with a larger work function, zinc, with $\phi \approx 4.3\mathrm{eV}$. It has the same slope $h$, so it runs parallel to the sodium line, but it starts at the larger threshold $f_0^{\prime }=4.3\mathrm{eV}/h\approx 1.04\times 10^{15}\mathrm{Hz}$, already in the ultraviolet.

The advanced view of the animation draws the sodium line with three points on it. The red point, at $4.28\times 10^{14}\mathrm{Hz}$, lies below the threshold $f_0=5.51\times 10^{14}\mathrm{Hz}$ and sits at zero: no electrons. The violet point, at $7.49\times 10^{14}\mathrm{Hz}$, lies at 0.82 eV, and the ultraviolet one, at $1.20\times 10^{15}\mathrm{Hz}$, at 2.68 eV: the two stopping potentials of Figure 1. The dim and bright violet runs share a single point, because intensity does not appear on this graph at all.

In the laboratory one measures $V_0$ rather than $K{E}_{\max }$. Dividing Einstein's equation by $e$ gives
$V_0=\frac{h}{e}f-\frac{\phi }{e}$
a straight line of slope $h/e=4.14\times 10^{-15}\mathrm{V}\cdot \mathrm{s}$: the stopping potential rises by about 0.41 V for every $10^{14}\mathrm{Hz}$. The slope is universal and only the intercept depends on the metal, so the lines for caesium, sodium and zinc are parallel, as in Figure 3. Millikan, who doubted Einstein's light quanta, measured such lines between 1914 and 1916 and found $h$ within about 1 % of today's value. Einstein's Nobel Prize for 1921 cited this law, not relativity.

Why a classical wave cannot do it

Light as a classical wave explains interference and diffraction perfectly, so why was a new idea needed at all? Because the wave picture gets all three observations of Figure 1 wrong.

The first is timing. A wave delivers energy continuously, spread over its whole front, and an electron can collect it only from roughly the area of its own atom, about $10^{-20}{\mathrm{m}}^2$. In dim light of 0.01 W/m² that area receives $0.01\cdot 10^{-20}=10^{-22}\mathrm{W}$, so gathering sodium's work function, $2.28\mathrm{eV}=3.65\times 10^{-19}\mathrm{J}$, would take
$t=\frac{3.65\times 10^{-19}\mathrm{J}}{10^{-22}\mathrm{W}}\approx 3650\mathrm{s}$
about an hour. Experiments find no delay at all, down to the few nanoseconds they can resolve.

The second is energy. Brighter light has a stronger electric field, which should shake the electrons harder and send them out faster; Figure 1 shows that brighter light gives more electrons, not faster ones. The third is the threshold. Given enough time, a wave of any frequency should deliver enough energy, yet below $f_0$ nothing comes out, however long one waits. All three predictions fail.

With photons all three facts follow at once. Absorption is all or nothing, so emission starts as soon as the first photon arrives. Each photon's energy $hf$ is fixed by the colour, so brighter light means more photons, not bigger ones. And a photon with less energy than $\phi$ frees nothing, however many of them arrive. A photon counter makes the graininess audible: dim the light and it does not click more quietly; its equal clicks just come less often.

A photon also carries momentum, $p=h/\lambda =E/c$, so $E^2-(pc{)}^2=0$: it is a particle of zero mass, as relativity requires of anything moving at $c$. Compton's scattering of X-rays off electrons confirmed this momentum in 1923. Where a particular photon will land can be predicted only as a probability, and that probability follows the intensity of the classical wave, which is why the wave picture still gets interference right.

A caution: a semiclassical model, in which the electrons obey quantum mechanics but the light stays a classical wave, also produces the threshold, the current proportional to intensity and the prompt emission. The photoelectric effect therefore points to photons without proving them. The proof came from experiments with single photons: antibunching in 1977, where light from a single atom was found never to deliver two photons at the same instant, and in 1986 the finding that one photon sent into a beam splitter never registers on both sides at once.

Beyond the single-photon threshold

The simple law assumes one photon per electron, a cold metal and an electron that must leave the material. Relaxing each assumption, or running the law backwards, gives four important extensions.

• Intense laser pulses deliver photons so densely that an electron can absorb $n$ of them at once: $K{E}_{\max }=nhf-\phi$. The rate grows as the $n$-th power of the intensity, so this needs very strong light. Two photons of the animation's red light, $2\cdot 1.77=3.54\mathrm{eV}$, would free electrons from sodium with up to 1.26 eV, more than single violet photons give.
• At any temperature above absolute zero a few electrons sit slightly above the Fermi level, so the threshold is not perfectly sharp but blurred over a few $kT$. Careful measurements of $\phi$ fit the shape of this thermal tail.
• Photoemission spectroscopy runs the equation backwards. With a known $hf$, usually ultraviolet or X-rays, the measured kinetic energy gives each electron's binding energy $E_{\mathrm{b}}$, its depth below the Fermi level (the $\epsilon$ of Figure 2), through $KE=hf-E_{\mathrm{b}}-\phi$.
• In a semiconductor a photon need not free the electron from the material, only lift it across the band gap, the energy gap between the filled states and the empty states in which an electron can carry current: 1.12 eV in silicon. The electron stays inside and adds to a current there. Silicon solar cells and camera sensors therefore respond to all wavelengths shorter than about 1100 nm.

Worked problem: measuring Planck's constant

The straight line of Figure 3 turns two stopping-potential readings into a measurement of $h$. Pretend you know neither $h$ nor $\phi$, only the animation's readouts: 0.82 V at 400 nm and 2.68 V at 250 nm.

First the frequencies. From $f=c/\lambda$ they are $7.495\times 10^{14}\mathrm{Hz}$ and $1.1992\times 10^{15}\mathrm{Hz}$, so $\Delta f=4.497\times 10^{14}\mathrm{Hz}$, while $\Delta V_0=2.68-0.82=1.86\mathrm{V}$.

Next write $e{V}_0=hf-\phi$ for each wavelength and subtract. The work function is the same in both equations, so it cancels, leaving $e\Delta V_0=h\Delta f$:
$h=\frac{e\Delta V_0}{\Delta f}=\frac{1.602\times 10^{-19}\cdot 1.86}{4.497\times 10^{14}}\approx 6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$

With $h$ known, either reading gives $\phi$. At 400 nm the photon energy is $hc/\lambda =1240/400=3.10\mathrm{eV}$, using $hc\approx 1240\mathrm{eV}\cdot \mathrm{nm}$, so $\phi =hf-e{V}_0=3.10-0.82=2.28\mathrm{eV}$. The line now predicts the stopping potential at any other wavelength, for instance $V_0=1240/300-2.28=1.85\mathrm{V}$ at 300 nm.

Now remove the collector and shine the ultraviolet light on an isolated sodium ball of radius $R=1.0\mathrm{cm}$. Every electron that escapes leaves the ball more positive, so each later electron must climb out of the ball's growing potential $V$, losing $eV$ on the way. Emission stops when $eV$ reaches $K{E}_{\max }$, that is at $V=2.68\mathrm{V}$: the ball charges itself up to the stopping potential. A sphere's potential is $V=Q/(4\pi {\epsilon }_{0}R)$, where ${\epsilon }_0$ is the permittivity of free space and $4\pi {\epsilon }_0=1.11\times 10^{-10}F/m$, so the final charge is
$Q=4\pi {\epsilon }_{0}RV=1.11\times 10^{-10}\cdot 0.010\cdot 2.68\approx 3.0\times 10^{-12}\mathrm{C}$
or $Q/e\approx 1.9\times 10^7$ missing electrons. Brighter light charges it faster, never higher.

Limits of validity

The clean law of Figure 3 rests on idealisations. Three matter in practice: the surface, the tube and the range of photon energies.

The work function belongs to a surface, not just to a metal: it changes with the crystal face exposed and even with a single layer of adsorbed atoms. Published values for sodium range from about 2.3 to 2.75 eV, and a caesium film can bring a surface below 1.5 eV. The yield is idealised too: the animation frees one electron for every photon above the threshold, whereas bare metals yield well under one electron per hundred photons, and the best photomultiplier cathodes about one per four.

A real tube usually has two different metals, the plate and the collector, and between two different metals joined in a circuit there is a contact potential, equal to the difference of their work functions divided by $e$. It shifts the measured cut-off: in an ideal tube, $e{V}_0$ actually equals $hf$ minus the collector's work function, not the plate's. The shift is the same at every frequency, so the slope $h/e$ is untouched, which makes the slope the robust result.

The energies here are tiny next to the electron's rest energy of 511 keV: the fastest ultraviolet electrons, with 2.68 eV, move at $9.7\times 10^{5}m/s$, 0.3 % of $c$, so $\frac{1}{2}m{v}^2$ is accurate. X-rays of tens of keV eject inner-shell electrons fast enough to need relativity. At higher energies other processes take over: above about 30 keV, Compton scattering overtakes photoabsorption in water and soft tissue, and above 1.022 MeV, twice the electron's rest energy, a photon can create an electron–positron pair.

Where this leads

Blackbody Radiation and the Quantum is where $h$ first appeared, in the spectrum of hot bodies; here it became a property of light itself. Atomic Models: the Bohr Atom applies $hf=E_2-E_1$ to spectral lines, Semiconductors, Diodes and LEDs runs the internal effect in reverse, with photons of energy close to the band gap, and Wave–Particle Duality follows single photons through a double slit and gives electrons the wavelength $h/p$. A photon's momentum ties the subject to Special Relativity and to Momentum and Collisions, and Particles and Forces: the Standard Model presents the photon as the carrier of the electromagnetic force.