In a school laboratory, two torches shine in turn into the same photocell. The blue torch is the dimmer of the two, yet it needs the larger retarding voltage to stop the current. The bright green torch gives the larger current, yet a small retarding voltage stops it. The meter and the voltmeter are telling two different stories about the same light.
One line for every metal
Start from Einstein's equation, \(eV_0 = h\nu - \phi\), and divide by \(e\):
\[ V_0 = \dfrac{h}{e}\,\nu - \dfrac{\phi}{e} \]
This has the form \(y = mx + c\), with \(y = V_0\) and \(x = \nu\). So a graph of the stopping potential against the frequency is a straight line. Its slope is \(h/e\), and its intercept on the \(V_0\) axis is \(-\phi/e\).
The slope contains only \(h\) and \(e\), two constants of nature. So the line has the same slope for every metal. The metal enters only through \(\phi\), which shifts the line sideways. The lines of different metals are therefore parallel. Each meets the frequency axis at its own threshold \(\nu_0 = \phi/h\).
Set sodium and a frequency of \(10 \times 10^{14}\,\text{Hz}\) in the sim. The photon energy is \(4.14\,\text{eV}\), and \(V_0 = 1.39\,\text{V}\). Now slide the frequency down. The stopping potential shrinks along the line and reaches zero at \(6.64 \times 10^{14}\,\text{Hz}\). Below that, the current vanishes at every \(V\). Try potassium, and the threshold moves to \(5.55 \times 10^{14}\,\text{Hz}\). Try copper, and it jumps to \(11.2 \times 10^{14}\,\text{Hz}\), in the ultraviolet.
Maximum kinetic energy against frequency
Some books plot \(K_{max}\) instead of \(V_0\). Then \(K_{max} = h\nu - \phi\), and the slope is \(h\) itself. The line meets the frequency axis at \(\nu_0\). Its dashed extension meets the \(K_{max}\) axis at \(-\phi\). The dashed part is only arithmetic: no electron ever has negative kinetic energy.
The \(K_{max}\) against \(\nu\) lines of two metals A and B are parallel. A meets the frequency axis at \(5 \times 10^{14}\,\text{Hz}\) and B at \(7 \times 10^{14}\,\text{Hz}\). Which statement is correct?
Two intensities, one frequency
Now look at the graph of the photocurrent \(I\) against the collector potential \(V\). Take two beams of the same colour, one twice as intense as the other. The brighter beam has twice as many photons each second. It frees twice as many electrons, so its saturation current is twice as high. Its photons are the same size, so both curves reach zero at the same stopping potential.
Two frequencies, one photon rate
Now take two beams of different frequency that deliver the same number of photons each second. They free the same number of electrons, so they saturate at the same current. But the higher frequency brings bigger photons and faster electrons. Its curve reaches zero further to the left, at a larger stopping potential.
Read the two chips of Fig 4.1 together. The height of a curve tells you how many electrons, and so how many photons, arrive each second. The point where it meets the \(V\) axis tells you how big each photon is. Intensity and frequency act on different parts of the graph.
Two beams of the same intensity but different colours do not send the same number of photons. The bluer beam has bigger photons, so it has fewer of them, and its saturation current is a little lower. When a question says the saturation currents are equal, it means the photon rates are equal.
Now the two torches of the laboratory make sense. The bright green torch sends more photons each second, so it frees more electrons and gives the larger current. The dim blue torch sends fewer photons, but each one is bigger. Its electrons leave faster, so it needs the larger retarding voltage to stop them. The meter counts photons; the voltmeter weighs each one.
Two I-V curves for one metal and one colour of light meet the \(V\) axis at the same point. Curve P saturates at \(6\,\mu\text{A}\) and curve Q at \(2\,\mu\text{A}\). What can you say?
Two beams on the same sodium surface, \(\phi = 2.75\,\text{eV}\), give the same saturation current. Beam A of \(310\,\text{nm}\) stops at \(-1.25\,\text{V}\). Beam B stops at \(-2.25\,\text{V}\). What is the wavelength of beam B, in nm?
Planck's constant from a voltmeter
The straight line is also a way to measure \(h\). Shine light of several known frequencies on one metal and measure \(V_0\) for each. Plot the points and draw the best straight line through them. Its slope is \(h/e\). The charge \(e\) is known, so \(h = e \times \text{slope}\).
In Fig 4.2 the best line passes through \(0.57\,\text{V}\) at \(8.0 \times 10^{14}\,\text{Hz}\) and \(2.22\,\text{V}\) at \(12.0 \times 10^{14}\,\text{Hz}\). Find Planck's constant, the threshold frequency and the work function.
Take two points on the line, far apart, not two data points.
\(\text{slope} = \Delta V_0 / \Delta\nu\), \(h = e \times \text{slope}\), \(\phi = h\nu_0\).
\(\text{slope} = (2.22 - 0.57)/(4.0 \times 10^{14}) = 4.125 \times 10^{-15}\,\text{V s}\). So \(h = 1.6 \times 10^{-19} \times 4.125 \times 10^{-15} = 6.60 \times 10^{-34}\,\text{J s}\). The line reaches \(V_0 = 0\) at \(\nu_0 = 8.0 \times 10^{14} - 0.57/4.125 \times 10^{-15} = 6.62 \times 10^{14}\,\text{Hz}\).
\(\phi = h\nu_0 = 6.60 \times 10^{-34} \times 6.62 \times 10^{14} = 4.37 \times 10^{-19}\,\text{J} = 2.73\,\text{eV}\), close to sodium's \(2.75\,\text{eV}\). A voltmeter and a set of filters have measured Planck's constant.
A student's best line of \(V_0\) against \(\nu\) passes through \(1.0\,\text{V}\) at \(7.2 \times 10^{14}\,\text{Hz}\) and \(3.0\,\text{V}\) at \(12.0 \times 10^{14}\,\text{Hz}\). What value of Planck's constant does it give? Answer in units of \(10^{-34}\,\text{J s}\).
For the same student's line, where does it meet the frequency axis, and what is the work function? Give the work function in eV.
Rank these metals by the frequency at which their \(V_0\) against \(\nu\) line meets the frequency axis, lowest first.
Robert Millikan of Chicago did not believe in Einstein's photons. From about 1905 he spent ten years testing the equation, hoping to find it wrong. The metal surfaces had to be perfectly clean. So he cut fresh surfaces of sodium, potassium and lithium with a knife worked from outside the vacuum. In 1916 he published the result: straight lines, with a slope that gave \(h = 6.57 \times 10^{-34}\,\text{J s}\), within one per cent of today's value. He had confirmed the equation he set out to overturn.
- Expecting a steeper \(V_0\) against \(\nu\) line for a metal with a larger work function.Every metal has the slope \(h/e\). The work function only shifts the line to a higher threshold.
- Reading the \(V_0\) axis intercept of the extension as a real stopping potential.The dashed extension below \(\nu_0\) is arithmetic. It gives \(-\phi/e\), not a measurement.
- Deciding the brighter beam from where the I-V curve meets the \(V\) axis.The height of the saturation current shows the intensity. The zero crossing shows the frequency.
- Computing the slope from two raw data points.Draw the best line first, then take two points on it, far apart.