§8 · photoelectric energy and counts

Letters: the paper takes the exciting light to consist of energy quanta of size $\frac{R}{N}\beta\nu$, writes P for the work an electron must do to leave the body (Φ here), and writes ε for the electron's charge. This explanation writes the quantum as hν, with h a later name for Rβ/N.

1. The simplest picture, which the paper assumes occurs: a light quantum gives its whole energy to a single electron. The paper also leaves open that some electrons take up only part of a quantum's energy.
2. An electron excited inside the body loses part of its kinetic energy before it reaches the surface. The fastest electrons are those excited right at the surface and moving straight out.
3. Each electron must do a work P, characteristic of the body, to leave it. So the kinetic energy of the fastest electrons is the quantum minus that work: $\frac{R}{N}\beta\nu-P$.
4. In the explanatory letters, with hν for the quantum and Φ for P, that is the first half of the formula below. Setting it to zero gives the lowest frequency at which the fastest electron can still escape, the threshold $\nu_0=\Phi/h$.

1. Worked case with modern SI constants, for a hypothetical work function of 2 electron volts at a frequency of 600 terahertz: $h\nu=6.62607\times10^{-34}\times6.00\times10^{14}=3.97564\times10^{-19}$ joule, which divided by $1.602177\times10^{-19}$ joule per electron volt is 2.4814 electron volts. Subtract 2: the maximum kinetic energy is 0.4814 electron volts.
2. The threshold for that work function is $\nu_0=2$ electron volts divided by h, about 484 terahertz.
3. Below the threshold, $h\nu-\Phi$ is negative. That is a shortfall of energy: no electron escapes by this route, and there is no negative kinetic energy to report.
4. The paper's own check, in its letters: charge the body to a positive potential Π that just prevents it from losing electricity. Then Π times the electron's charge ε equals the kinetic energy $\frac{R}{N}\beta\nu-P$.
5. Multiply by N: the charge becomes E, the charge of one gram-equivalent of a monovalent ion, and the right side becomes Rβν minus the escape work for that amount of charge. The paper sets that work to zero for an estimate, takes $E=9{,}6\cdot10^{3}$ in electromagnetic units, $\nu=1{,}03\cdot10^{15}$, the limit of the solar spectrum toward the ultraviolet, and $\beta=4{,}866\cdot10^{-11}$.
6. Then Π = Rβν/E: $8.31\times10^{7}\times4.866\times10^{-11}\times1.03\times10^{15}/(9.6\times10^{3})=4.34\times10^{8}$ electromagnetic units, with R the standard value of 1905. One volt is $10^{8}$ of those units, so Π is about 4.34 volts. The page gives 4,3 volts and says the result agrees in order of magnitude with Lenard's.
7. The paper's further prediction: Π, plotted against the frequency of the exciting light, must be a straight line whose slope does not depend on the substance. In the explanatory letters the slope of $K_{\mathrm{max}}$ against ν is h for every metal, and only the intercept, set by Φ, changes.
8. Power sets the count, not the energy of each electron. Optical power is energy per second; divided by hν it gives quanta per second. At 600 terahertz, one milliwatt carries $2.5\times10^{15}$ quanta per second, and two milliwatts carry twice as many, each still of 2.4814 electron volts.
9. So, if each quantum gives up its energy independently of all the others, the speeds of the electrons do not depend on the intensity of the light, while the number leaving the body is proportional to it, other things being equal (p. 147). Twice the power can give twice the electrons per second; it cannot raise the maximum energy of one.
10. The laboratory's straight line follows from the equation it is programmed with. It shows a consequence of the model; only measurements made independently can test the model.

Letters: Π is the potential the illuminated body reaches, E the charge of a gram-equivalent of a univalent ion, $P'$ the work for a gram-equivalent of electrons to leave the body, R the gas constant, N the number of real molecules in a gram-equivalent, β the constant from Planck's formula in §2, and ν the frequency. (R/N)βν is the energy of one quantum; later physics writes it hν.

1. The regularities being qualified come from the paragraph before: each quantum gives its energy to electrons independently of all the others.
2. Consequence one: the speeds of the escaping electrons, the quality of the cathode rays, do not depend on the intensity of the light, because each electron still receives the energy of a single quantum.
3. Consequence two: the number of electrons leaving is proportional to the intensity, other things equal, because doubling the intensity doubles the number of quanta arriving.
4. Both rest on one quantum for one electron, which gives the stopping condition, for a gram-equivalent of electrons:

1. Worked case, with the paper's own check: $P'=0$, $R=8.31\times10^{7}$, $\beta=4.866\times10^{-11}$, $\nu=1.03\times10^{15}$ and $E=9.6\times10^{3}$. Then $R\beta\nu=8.31\times10^{7}\times4.866\times10^{-11}\times1.03\times10^{15}\approx4.16\times10^{12}$, and $Π=R\beta\nu/E\approx4.3\times10^{8}$, which the paper converts to about 4.3 volts. One quantum, one electron.
2. The first condition from §7, carried over: if so many quanta are being transformed at once in each unit of volume that one electron could take energy from two quanta, the energy available to it could reach twice (R/N)βν, and the equation above would no longer bound it. The paper names no intensity at which that begins.
3. The second condition from §7, carried over: the quantum picture came from the entropy of radiation within the range of Wien's law (§6). Light well outside that range, such as light from a very hot source at the wavelength used, is not covered by that argument, so the argument does not fix the size of its energy portions.
4. What Einstein asserts is only that such remarks would have to be made. He calls the limits presumable and leaves their details open, so neither carried-over case is a prediction of the paper.

Letters: the paper charges the illuminated body to a positive potential Π and writes ε for the electron's charge. This explanation writes $V_s$ for the size of the stopping potential and e for the positive size of the electron's charge; the electron's charge itself is −e.

1. Set the arrangement and the sign: the body is at the potential $V_s$ above the conductors around it, which are at zero. An electron leaving the body is pulled back toward it, so the potential retards the electron.
2. An electron moving from the body out to the surroundings, across a potential difference $V_s$, loses kinetic energy $eV_s$. If it had kinetic energy K at the surface, it arrives with $K-eV_s$, or is turned back if that would be negative.
3. The body stops losing electricity when even the fastest electron is turned back, that is, when $eV_s$ equals the largest kinetic energy $K_{\mathrm{max}}$.
4. Insert the one-quantum budget of the previous passage for $K_{\mathrm{max}}$.

1. Divide by e: $V_s=(h\nu-\Phi)/e$. Worked with the previous passage's hypothetical numbers: $K_{\mathrm{max}}$ is 0.4814 electron volts, so $V_s$ is 0.4814 volts, since one electron volt is e times one volt.
2. Keep sizes and signs apart. $V_s$ is a size, zero or positive. The electron's charge is negative, and writing e for its size does not change that. A voltage read from an instrument also needs its reference electrode named.
3. The paper's gram-equivalent form of the stopping condition is $ΠE=R\beta\nu-P'$, with $P'$ the potential of a gram-equivalent of negative electricity with respect to the body: the work for that much electricity to leave it. A footnote attached to it takes up one more possibility.
4. Suppose the light must first free each electron from a neutral molecule, at the cost of a certain work, before the electron can make its way out. The footnote says nothing in the derived relation changes; $P'$ is then to be read as a sum of two terms.
5. Why nothing changes: one quantum's energy, (R/N)βν per electron ($h\nu$ in the explanatory letters), now pays two works in turn, freeing the electron from its molecule and taking it out of the body. Both come out of the same quantum, so the fastest electron keeps the quantum's energy less their sum. Write $\Phi$ for the sum, and $eV_s=h\nu-\Phi$ stands as before.
6. Worked case, with hypothetical numbers: freeing the electron costs 1.0 electron volt and leaving the body 2.0, so $\Phi$ is 3.0 electron volts. A quantum of 3.5 electron volts leaves the fastest electron 3.5 − 1.0 − 2.0 = 0.5 electron volts, and $V_s$ is 0.5 volts. A quantum of 2.5 electron volts releases no electron from the body.
7. What stays: the frequency enters only through the quantum's energy, so the stopping potential plotted against frequency is still a straight line whose slope does not depend on the substance. The two works together only move the frequency at which the line reaches zero.
8. Partial transfer and losses: if an electron receives only part of a quantum, or loses energy inside the body, it reaches the surface with less than $h\nu-\Phi$. The paper then replaces the equation by an inequality: in its letters, Π times E, plus the escape work for a gram-equivalent, is at most Rβν.
9. In the explanatory letters: $eV_s\le h\nu-\Phi$. Losses can only lower the stopping potential; they cannot give one electron more than the quantum it received.
10. The inverse process, cathode luminescence: electrons that have fallen through a potential produce light. Reversing the budget, the energy supplied per electron must be at least the energy of the quantum produced, $eV_{\mathrm{acc}}\ge h\nu$ in the explanatory letters.
11. Worked scale from the paper: in the substances Lenard examined, the cathode rays had to fall through several hundred, in some cases several thousand, volts to produce visible light, far more than the few volts one quantum of visible light needs. The paper concludes that the kinetic energy of one electron is used to produce many light quanta.
12. A bound is not a promise: $eV_{\mathrm{acc}}\ge h\nu$ states what is needed in energy, not that every collision produces light.