§9 · ionization bounds and closing scope

Letters: the paper writes J for the work needed to ionize one gram-equivalent, L for the amount of light absorbed (not the speed of light here), and j for the number of gram-molecules ionized. This explanation works per molecule: I is the energy needed to ionize one molecule, hν is one quantum, $E_{\mathrm{abs}}$ is the absorbed energy, and $n_{\mathrm{ions}}$ is the count of ions.

1. The assumption: when a gas is ionized by ultraviolet light, each absorbed quantum is used to ionize one gas molecule.
2. A single quantum can pay the cost of ionization only if it carries at least that much energy. Per molecule, that is the formula below.

1. The paper's form, for one gram-equivalent: multiply both sides by N, and the condition reads Rβν at least J.
2. Worked with the paper's inputs. Lenard's longest wavelength that still ionizes air is about $1{,}9\cdot10^{-5}$ cm. Its frequency is the speed of light divided by the wavelength, $3\times10^{10}/(1.9\times10^{-5})=1.58\times10^{15}$ per second. Then Rβν is $8.31\times10^{7}\times4.866\times10^{-11}\times1.58\times10^{15}=6.4\times10^{12}$ erg, with R the standard value of 1905; the page prints $6{,}4\cdot10^{12}$ erg.
3. So J is at most about $6.4\times10^{12}$ erg per gram-equivalent. An independent upper bound comes from Stark's smallest measured ionizing voltage for air, about 10 volts: 10 volts times the charge of a gram-equivalent, $9.6\times10^{3}$ electromagnetic units, is $10\times10^{8}\times9.6\times10^{3}=9.6\times10^{12}$ erg. The paper remarks that the two bounds are nearly equal.
4. The footnote on Stark's figure: inside the gas, the ionizing voltage for negative ions is, to be sure, five times larger. The 10 volts is the smallest value measured, at platinum anodes.
5. Why the smallest value is the one to use: the paper takes an ionizing voltage as giving an upper limit. A charge that has fallen through it had that much energy available to ionize a molecule, so the work of ionization cannot be larger. Each measured voltage gives a cap, and the smallest gives the tightest.
6. Worked with the footnote's factor: five times 10 volts is 50 volts, and $50\times10^{8}\times9.6\times10^{3}=4.8\times10^{13}$ erg per gram-equivalent, 7.5 times Lenard's figure. That is a looser upper limit, not a contradiction; the near agreement the paper notes is between Lenard's figure and the smallest voltage.
7. Now the count. With light of one frequency, the absorbed energy divided by the energy of one quantum is the number of quanta absorbed.
8. If each absorbed quantum ionizes at most one molecule, the number of ions cannot exceed that number of quanta.

1. The paper's stronger statement assumes exactly one ionization for each absorbed quantum: $j=L/(R\beta\nu)$ gram-molecules for absorbed light L. It states the condition: the gas must show, at that frequency, no noticeable absorption that is not accompanied by ionization.
2. Worked count: absorb 1 erg of light at Lenard's limit. Per molecule, one quantum carries $6.4\times10^{12}/(6.17\times10^{23})=1.03\times10^{-11}$ erg, so 1 erg is about $9.7\times10^{10}$ quanta, and at most that many ions.
3. Meeting the threshold makes an ionization possible, not certain. An unknown absorbed fraction or yield stays unknown rather than being set to 100 percent.
4. The paper calls a test of this count by experiment very important, and ends there, dated Bern, 17 March 1905.