On a winter evening you can sit beside an electric heater for an hour. Its glow pours energy onto your skin, and your skin gets warm, but it never burns. Twenty minutes of noon sun on a beach at Goa in May can leave the same skin red and sore the next day. The heater can deliver as much energy per square centimetre. Something other than the amount of energy is at work.
Light as a wave
By 1900, light was known to be a wave. Its interference and diffraction had been measured with great care, as the chapter on interference shows. In a wave, energy spreads smoothly over the whole wavefront and arrives continuously. The intensity of the wave, its power per square metre, grows with the square of its amplitude. A brighter light is a wave of larger amplitude.
In this picture an electron in the metal is shaken by the wave. It gathers energy steadily, like a swing pushed in time, until it has enough to escape. Follow that idea through, and it makes three clear predictions about the photoelectric effect.
Three predictions, three failures
Intensity. A brighter wave shakes the electron harder. So brighter light should free faster electrons, and the stopping potential should grow with intensity. The photocell says otherwise: \(V_0\) does not move when the intensity changes.
Frequency. A wave of any frequency carries energy. Make red light bright enough, or wait long enough, and it should free electrons too. Instead there is a sharp threshold frequency. Below it, no intensity and no wait frees a single electron.
Time. In dim light an electron receives only a trickle of energy. It should need seconds, minutes or hours to gather \(\phi\). Instead the first electrons leave within a nanosecond, whatever the intensity.
A faint lamp gives an intensity of \(0.01\,\text{W/m}^2\) on a sodium plate, with \(\phi = 2.75\,\text{eV}\). In the wave picture, suppose an electron gathers the energy falling on one atom, a patch about \(0.1\,\text{nm}\) by \(0.1\,\text{nm}\). Estimate how long it waits to gather \(\phi\), in seconds.
According to the wave theory of light, what should happen to the photoelectrons when the intensity of the light is doubled?
Einstein's packets of light
In 1900 Max Planck had explained the colours of glowing hot bodies with a strange assumption. The atoms of the hot body could give out light energy only in steps of size \(h\nu\). Here \(\nu\) is the frequency and \(h\) is a new constant of nature, now called Planck's constant. Planck treated the steps as a trick of the emitters, not a property of light.
In 1905 Albert Einstein went much further. He proposed that light itself travels as packets of energy, later named photons. Each packet carries energy \(h\nu\). When light is absorbed, a whole photon is absorbed by one electron, in one event. A photon is never split and never shared.
In this picture the intensity of light means something new. It is the number of photons crossing each square metre per second, times the energy \(h\nu\) of each. Brighter light of one colour has more photons, not bigger ones. The size of each photon is set by the colour alone.
Find the energy of a photon of frequency \(6.0 \times 10^{14}\,\text{Hz}\), in joules and in eV. Then find the energy of a violet photon of wavelength \(400\,\text{nm}\).
(a) \(E = h\nu = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19}\,\text{J}\).
(b) In eV, \(E = 3.98 \times 10^{-19} / 1.6 \times 10^{-19} = 2.49\,\text{eV}\). This is green light.
(c) For \(\lambda = 400\,\text{nm}\), \(E = 1240/400 = 3.10\,\text{eV}\), with no powers of ten at all.
Values used: \(h = 6.63 \times 10^{-34}\,\text{J s}\), \(e = 1.6 \times 10^{-19}\,\text{C}\), \(hc = 1240\,\text{eV nm}\).
Find the energy of one photon of frequency \(5.0 \times 10^{14}\,\text{Hz}\). Give the answer in units of \(10^{-19}\,\text{J}\).
The photoelectric equation
Now follow one photon into the metal. It gives all its energy \(h\nu\) to one electron. The electron pays at least the work function \(\phi\) to leave the surface. Whatever is left becomes kinetic energy. The most loosely held electrons pay exactly \(\phi\), so they leave with the most, \(K_{max}\). Deeper electrons lose more on the way out and leave slower.
\[ h\nu = \phi + K_{max} \]
Why. energy is conserved in the single event. The photon's energy pays the work function of the loosest electron, and the rest is kinetic energy.
\[ K_{max} = h\nu - \phi \]
Why. rearrange; any other electron pays more than \(\phi\) and leaves with less.
\[ K_{max} = eV_0 \]
Why. the stopping potential measures the energy of the fastest electron (lesson 2).
\[ eV_0 = h\nu - \phi \]
Read it. put the two expressions for \(K_{max}\) together. The measured \(V_0\) is tied to the frequency and the metal.
A photon with less energy than \(\phi\) cannot free an electron at all. Emission therefore needs \(h\nu \ge \phi\). The threshold frequency is \(\nu_0 = \phi/h\), and the matching threshold wavelength is \(\lambda_0 = hc/\phi\). Light of longer wavelength than \(\lambda_0\) frees nothing, however bright. For sodium, \(\lambda_0 = 1240/2.75 = 451\,\text{nm}\); for caesium, \(1240/2.14 = 579\,\text{nm}\).
Green light of \(520\,\text{nm}\) frees electrons from a caesium plate. The same green light is made twice as bright. What happens to the freed electrons?
Now try the other chips of the figure. Violet light of \(400\,\text{nm}\) brings bigger photons, \(3.10\,\text{eV}\) against \(2.38\,\text{eV}\). The electrons leave with up to \(0.96\,\text{eV}\) instead of \(0.24\,\text{eV}\), about four times the energy and twice the speed. Bright red light of \(650\,\text{nm}\) brings many photons of \(1.91\,\text{eV}\) each. Every one is below the \(2.14\,\text{eV}\) that caesium demands, so none frees an electron.
The three failures, answered
The photon picture now explains each fact of the photocell in turn.
- Intensity. Brighter light means more photons of the same size. More photons free more electrons, so the current grows. Each electron still receives \(h\nu\), so \(K_{max}\) and \(V_0\) do not change.
- Threshold. One photon frees one electron, and photons do not team up. If \(h\nu < \phi\), no single photon can free an electron, and a billion of them fail a billion times.
- No delay. The energy arrives in one packet, not as a slow trickle. The first photon that reaches an electron can free it at once.
Violet light of wavelength \(400\,\text{nm}\) falls on sodium, with \(\phi = 2.75\,\text{eV}\). Find the photon energy, the largest kinetic energy of the photoelectrons, the stopping potential and the threshold wavelength.
Work in eV and nm, with \(hc = 1240\,\text{eV nm}\).
\(E = 1240/\lambda\), \(K_{max} = E - \phi\), \(V_0 = K_{max}/e\), \(\lambda_0 = 1240/\phi\).
\(E = 1240/400 = 3.10\,\text{eV}\). \(K_{max} = 3.10 - 2.75 = 0.35\,\text{eV}\). So \(V_0 = 0.35\,\text{V}\). \(\lambda_0 = 1240/2.75 = 451\,\text{nm}\).
Most of each photon's energy, \(2.75\,\text{eV}\) of \(3.10\,\text{eV}\), pays the work function. The electrons keep only \(0.35\,\text{eV}\).
Light of wavelength \(500\,\text{nm}\) falls on potassium, with \(\phi = 2.30\,\text{eV}\). What is the stopping potential, in volts?
Light of wavelength \(500\,\text{nm}\) falls in turn on four metals. Which of them give photoelectrons?
For one metal, light of \(248\,\text{nm}\) gives a stopping potential of \(2.25\,\text{V}\), and light of \(310\,\text{nm}\) gives \(1.25\,\text{V}\). What is the work function of the metal, in eV?
Light of \(400\,\text{nm}\) frees electrons from sodium with \(K_{max} = 0.35\,\text{eV}\). The wavelength is halved to \(200\,\text{nm}\). What is \(K_{max}\) now?
The arc, the lamp and the heater
The welder's arc of lesson 1 gives out ultraviolet light, with photons of more than \(4\,\text{eV}\). Zinc has a work function of about \(4.3\,\text{eV}\), so its threshold wavelength is \(1240/4.3 = 288\,\text{nm}\). The arc's shortest-wavelength photons can free electrons from zinc. A red stage lamp sends photons of about \(1.9\,\text{eV}\). However many it sends, not one of them can pay \(4.3\,\text{eV}\).
Your skin works the same way. Ultraviolet photons in sunlight, near \(300\,\text{nm}\), carry about \(4.1\,\text{eV}\) each. That is enough for a single photon to break a molecule in a skin cell. The heater's infrared photons carry less than \(1\,\text{eV}\) each. They can warm the skin, but none of them can break those molecules.
In March 1905 Albert Einstein, then a clerk at the patent office in Bern, sent a paper to the journal Annalen der Physik. Its title called the light quantum only "a heuristic point of view", a useful guess. Few physicists believed it for years, since the wave nature of light seemed settled. When Einstein received the Nobel Prize in Physics for 1921, the citation named "the law of the photoelectric effect", not relativity. The law was the equation of this lesson.
- Adding the work function to the photon energy.The work function is a cost paid on the way out: \(K_{max} = h\nu - \phi\).
- Expecting \(K_{max}\) to double when the frequency doubles.The photon energy doubles; \(K_{max}\) grows by more, since \(\phi\) is subtracted first.
- Thinking enough photons below the threshold can together free an electron.One photon frees one electron. Photons below \(h\nu = \phi\) fail one by one.
- Using \(\lambda\) in metres in \(E = 1240/\lambda\).The shortcut needs \(\lambda\) in nanometres and gives \(E\) in eV.