Light arrives in individual events
Shine a dim blue lamp on a clean piece of sodium in a vacuum and electrons fly off the metal at once. Replace it with a powerful red floodlight, thousands of times brighter, and nothing comes off at all, however long you wait. Why does the colour of the light matter, and the brightness not?
The previous essay ended with a question: Planck's energy steps described how matter and radiation trade energy inside a hot box, but did they say anything about light itself? This essay answers yes. Two experiments, one with metals and one with X-rays, show light delivering its energy and its momentum in separate packets. Along the way we will compute the energy budget of a single electron leaving sodium, and see why one of the two effects shows up only with X-rays.
What a wave of light should do
In 1900 light was, beyond reasonable doubt, a wave of electric and magnetic fields. That picture explained interference, diffraction and polarisation, and it made firm predictions about what happens when such a wave washes over a metal.
The oscillating electric field shakes the electrons in the surface. A brighter wave has a stronger field, shakes them harder, and should send them out faster. A dimmer wave should still get electrons out eventually: an electron keeps absorbing energy from the wave until it has enough to escape, so dim light should mean a delay before anything happens, not a refusal. And frequency should play no special role, since energy delivered is energy delivered.
The experiment, the photoelectric effect, contradicts all three. Light falls on a metal surface in a vacuum and a nearby electrode collects the electrons that come off. Two quantities can be measured: the current, which counts electrons per second, and the energy of the fastest electrons, found by making the collecting electrode negative and raising that voltage until even the fastest electrons are turned back. The voltage that just stops the current is the stopping potential, and in electronvolts it equals the maximum kinetic energy $K_{\max}$ of the electrons.
- Brightness changes the number, not the energy. Doubling the intensity doubles the current. $K_{\max}$ does not change at all.
- Frequency changes the energy. Raise the frequency and $K_{\max}$ rises with it.
- There is a threshold. Below a certain frequency, which depends on the metal, no electrons come out at any intensity. Above it they come out immediately, even in very dim light.
Einstein's proposal: light comes in quanta
In 1905 Albert Einstein proposed an explanation that goes further than Planck had. Light itself, he suggested, consists of separate quanta, each carrying energy $h\nu$. Today they are called photons. A photon is absorbed whole, by a single electron, or not at all.
To finish the account we need one fact about metals. Their outer electrons move freely inside the metal but are held in at its surface: pulling one out costs a minimum energy, called the work function $W$. Its measured value depends on how clean the surface is: for sodium, reported values run from about 2.3 to 2.75 eV, and this essay uses 2.3 eV, an often-quoted figure; for zinc it is about 4.3 eV. The electron that absorbs a photon spends at least $W$ of its energy getting out of the metal, and whatever is left is its kinetic energy. The fastest electrons are those that lose nothing else on the way out:
$$K_{\max} = h\nu - W .$$Each observation now follows. A brighter beam of the same colour carries more photons per second, each with the same energy $h\nu$: more electrons, none of them faster. Raising the frequency raises the energy of every photon. And if $h\nu < W$, no single photon can free an electron, and because an electron cannot collect several photons at once in ordinary light, piling on more of them does not help. The threshold frequency is $\nu_0 = W/h$.
The equation also makes a prediction beyond the observations it was built to explain. Plot $K_{\max}$ against $\nu$ and you should get a straight line whose slope is $h$, the same for every metal, while the work function only shifts the line sideways. Robert Millikan, who expected to refute Einstein, tested this carefully and published the result in 1916: straight lines, and a slope matching Planck's constant from the blackbody spectrum. (He confirmed the equation while remaining unconvinced for some years that light itself comes in quanta.) A number first obtained from the glow of a hot box reappeared in electrons leaving a metal.
Worked example: sodium under red and blue light
A shortcut. Photon energies come up constantly, so it pays to combine the constants once: $hc = (6.626\times10^{-34}\ \text{J s})(2.998\times10^{8}\ \text{m/s}) = 1.986\times10^{-25}$ J m, which in electronvolts and nanometres is $hc = 1240$ eV nm. Then a photon's energy in eV is 1240 divided by its wavelength in nm.
Red light, 700 nm. One photon carries
$$h\nu = \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{700\ \text{nm}} = 1.77\ \text{eV}.$$That is less than sodium's 2.3 eV work function. No photon of this light can free an electron, so the red floodlight ejects nothing, whatever its brightness.
Blue light, 450 nm. Now $h\nu = 1240/450 = 2.76$ eV, and
$$K_{\max} = 2.76\ \text{eV} - 2.3\ \text{eV} = 0.46\ \text{eV}.$$A collecting electrode held 0.46 V negative will just stop the fastest electrons. With $K = \tfrac12 m v^2$ and $m = 9.11\times10^{-31}$ kg, they leave at about $4\times10^{5}$ m/s.
The threshold. Sodium's cut-off is where $h\nu = W$: $\lambda_0 = 1240/2.3 = 540$ nm, in the green. Anything bluer ejects electrons; anything redder does not. (With the higher reported value, 2.75 eV, the threshold moves to 450 nm and our blue light would only just manage it; the argument about colour and brightness is the same either way.)
Check: how many photons? A 1 mW red laser pointer at 650 nm sends out $10^{-3}$ J each second in photons of $1.91\ \text{eV} = 3.06\times10^{-19}$ J, which is $3\times10^{15}$ photons per second. Ordinary light is a torrent of photons, which is why it looks continuous. The graininess only matters when a single photon has to do a job on its own, as in freeing one electron.
Optional: the waiting time a wave would need
The wave picture says a dim light should eventually eject electrons, after a delay while energy accumulates. One can estimate that delay. Take light of intensity $I$ (watts per square metre) and suppose, generously, that an atom at the surface collects all the energy falling on an area the size of the atom, about $10^{-19}$ m². At a feeble $10^{-6}$ W/m², it collects $10^{-25}$ J per second, and needs $W \approx 4\times10^{-19}$ J: a wait of more than a month.
Experiments find no such delay. Even in very dim light, some electrons emerge within a few billionths of a second of the light being switched on, just as if the energy arrived in one lump at one place. Born's Atomic Physics describes an early test of exactly this kind by Meyer and Gerlach (1914), with tiny suspended specks of metal whose charge could be watched as they lost electrons.
Photons also carry momentum
The photoelectric effect shows light delivering energy in lumps. A particle should also carry momentum, and a light wave was already known to push: classical electromagnetism gives a beam carrying energy $E$ a momentum $E/c$. Applied to a single photon of energy $h\nu$, that is
$$p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}.$$This follows too from the relation between energy and momentum in special relativity, $E^2 = p^2c^2 + m^2c^4$, which the optional essays on relativity derive. Here we only need it stated. For an object with no mass, $m = 0$, it gives $E = pc$ exactly.
In the early 1920s Arthur Compton scattered X-rays off the loosely held electrons in light materials such as graphite and measured the wavelength of the scattered rays. The wave picture predicts no change: an electron shaken by a wave re-radiates at the same frequency. Compton found the scattered X-rays had a longer wavelength, by an amount that grew with the scattering angle $\theta$ and did not depend on the original wavelength at all.
Treat the scattering as a collision between a photon and an electron at rest, with energy and momentum both conserved, and the result is
$$\lambda' - \lambda = \frac{h}{m_e c}\,(1 - \cos\theta).$$The photon hands some of its energy to the recoiling electron, so it leaves with less energy and therefore a longer wavelength. The combination $h/m_ec = 2.43\times10^{-12}$ m is called the Compton wavelength of the electron. Two checks show the formula is sensible. At $\theta = 0$ the photon carries straight on, nothing has been transferred, and the shift is zero. At $\theta = 180°$ the photon bounces straight back, the electron takes the largest possible kick, and the shift is largest, $2h/m_ec = 4.85$ pm. Compton measured just this dependence, and the recoiling electrons were seen too.
Optional: deriving the Compton shift
Before the collision: a photon of momentum $p = h/\lambda$ and energy $pc$, and an electron at rest with energy $m_ec^2$. After: a photon with momentum $p'$ at angle $\theta$, and an electron with momentum $\vec P$ and energy $\sqrt{P^2c^2 + m_e^2c^4}$ (the relativistic relation again; the recoiling electron can be fast).
Momentum conservation, $\vec p = \vec p\,' + \vec P$, gives the electron's momentum as $\vec P = \vec p - \vec p\,'$. Squaring,
$$P^2 = p^2 + p'^2 - 2pp'\cos\theta .$$Energy conservation, $pc + m_ec^2 = p'c + \sqrt{P^2c^2 + m_e^2c^4}$. Move $p'c$ to the left, square both sides, and cancel $m_e^2c^4$:
$$(p - p')^2 c^2 + 2(p - p')\,m_ec^3 = P^2c^2 .$$Substitute $P^2$ from the momentum equation. The $p^2$ and $p'^2$ terms cancel, leaving $(p - p')\,m_ec = pp'(1 - \cos\theta)$. Divide by $pp'm_ec$:
$$\frac{1}{p'} - \frac{1}{p} = \frac{1 - \cos\theta}{m_ec}.$$Since $1/p = \lambda/h$, multiplying by $h$ gives the Compton formula.
Running the idea backwards
By the mid-1920s light had a double identity. It spreads, diffracts and interferes like a wave, with a wavelength $\lambda$. It delivers energy and momentum in single events, like a particle, with the two descriptions tied by $p = h/\lambda$.
In 1924 Louis de Broglie proposed reading the same relation in the other direction. If a wave of light has particle properties, perhaps a particle of matter, an electron, has wave properties, with a wavelength
$$\lambda = \frac{h}{p}.$$For everyday objects this wavelength is absurdly small, far too small ever to notice. For an electron with a few tens of electronvolts of energy it is a fraction of a nanometre, about the spacing of atoms in a crystal. In 1927 electrons reflected from crystals were found to form diffraction patterns, at the angles this wavelength predicts. The double identity is not special to light. We will lean on this relation heavily from the double-slit essay onward.
Two questions
1. Zinc in ultraviolet. Zinc has a work function of about 4.3 eV. Does ultraviolet light of 300 nm eject electrons from it? What about 250 nm, and what stopping potential would you measure then?
Worked answer
At 300 nm, $h\nu = 1240/300 = 4.13$ eV, just short of 4.3 eV. No electrons come out, whatever the intensity. At 250 nm, $h\nu = 1240/250 = 4.96$ eV, so $K_{\max} = 4.96 - 4.3 = 0.66$ eV, and the stopping potential is 0.66 V.
Zinc's threshold is $1240/4.3 = 288$ nm. So 300 nm light, which is ultraviolet and invisible to the eye, still fails. "Ultraviolet" is not a guarantee; the comparison is always with the particular metal's work function.
2. Why X-rays? The Compton shift at 90° is 2.4 pm whatever the wavelength of the light. Why, then, was it discovered with X-rays of about 0.07 nm, and never noticed with visible light? Does visible light not carry momentum?
Worked answer
It does, and the shift is the same 2.4 pm. What differs is the shift as a fraction of the wavelength. For a 71 pm X-ray, 2.4 pm is 3.4 percent, easily measured. For 500 nm green light it is about 5 parts in a million, buried in the natural width of any real spectral line.
The physical reason: a visible photon's momentum, $h/\lambda \approx 10^{-27}$ kg m/s, is far too small to make an electron recoil noticeably. An X-ray photon carries about 7,000 times more and can knock the electron away with hundreds of electronvolts. The tempting wrong turn is to read "the shift doesn't depend on wavelength" as "the effect is equally visible at every wavelength". Whether a change is noticeable depends on its size relative to what it changes.
Further reading
- Born, Atomic Physics (Dover, 8th ed.), Ch. IV §2 "Light Quanta" and §4 "Compton Effect", with the full Compton calculation in Appendix X. A clear account of the experiments, including the Meyer–Gerlach metal-dust test and the coincidence measurements of Bothe and Geiger.
- Feynman, Lectures on Physics, Vol. I, Ch. 34 "Relativistic Effects in Radiation", §34-9 "The momentum of light": why a beam of light pushes, and the momentum of a photon.
Where this leaves us
Colour matters and brightness does not because light delivers its energy to an electron in whole photons of $h\nu$. Brightness sets how many photons arrive; colour sets what each one can do. Below sodium's threshold in the green, no single photon can pay the 2.3 eV exit fee, and a thousand of them arriving separately cannot pool their money.
If light exchanges energy in fixed amounts, what about the atoms that emit and absorb it? A hot gas of hydrogen does not glow with a smooth blackbody spectrum. It emits a handful of sharp colours, and nothing in between, as if each atom could only release energy in a few particular amounts. The next essay follows the first serious attempt to explain why, a model that got the colours right for reasons that turned out to be wrong.