§9 · ionization bounds and closing scope · Light quanta: from entropy to an energy scale

§9 · ionization bounds and closing scope

A threshold does not specify a yield

What can energy conservation say about gas ionization?

Let $I$ be the required energy for the specified ionization channel. A single quantum can pay this cost only when $h\nu$ is at least $I$ . A smaller quantum leaves an energy deficit; a larger quantum makes the event energetically possible, not certain.

h\nu\ge I

One-quantum ionization requires quantum energy at least equal to the specified ionization energy.

For monochromatic absorbed energy $E_\text{abs}$ , the number of absorbed quanta is represented by $E_\text{abs}/(h\nu)$ . With at most one counted ionization per quantum this gives an upper bound. It becomes an equality only when all absorbed quanta each produce one counted ionization.

n_{\mathrm{ions}}\le\frac{E_{\mathrm{abs}}}{h\nu}

The ionization count is at most absorbed energy divided by quantum energy, under the at-most-one-ionization assumption.

The paper compares two upper limits for the work of ionizing a gram-equivalent of air. Lenard's longest ionizing wavelength gives $6.4\times10^{12}$ erg, and Stark's smallest measured ionizing voltage, about 10 volts at platinum anodes, gives $9.6\times10^{12}$ , nearly the same. A footnote concedes that inside the gas the ionizing voltage for negative ions is five times larger. Each measured voltage caps the work, so the smallest gives the tightest cap; the fivefold figure caps it only at about $4.8\times10^{13}$ erg, and the near agreement the paper notes holds for the smallest value.

The ionization laboratory separates incident energy, absorbed fraction, threshold, and conversion assumptions. Its historical checks use their own stated units; this explanation does not infer a new gas species or a real-material yield from a threshold slider.

Return to the energy threshold and the separately conditioned count bound. (included in this file)

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.

The assumption: when a gas is ionized by ultraviolet light, each absorbed quantum is used to ionize one gas molecule.
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.

h\nu\ge I

One-quantum ionization requires quantum energy at least equal to the specified ionization energy.

The paper's form, for one gram-equivalent: multiply both sides by N, and the condition reads Rβν at least J.
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.
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.
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.
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.
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.
Now the count. With light of one frequency, the absorbed energy divided by the energy of one quantum is the number of quanta absorbed.
If each absorbed quantum ionizes at most one molecule, the number of ions cannot exceed that number of quanta.

n_{\mathrm{ions}}\le\frac{E_{\mathrm{abs}}}{h\nu}

The ionization count is at most absorbed energy divided by quantum energy, under the at-most-one-ionization assumption.

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.
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.
Meeting the threshold makes an ionization possible, not certain. An unknown absorbed fraction or yield stays unknown rather than being set to 100 percent.
The paper calls a test of this count by experiment very important, and ends there, dated Bern, 17 March 1905.

Return to the energy threshold and the separately conditioned count bound. (included in this file)

No more ions than quanta

Modern teaching equation, not the equation as printed.

n_{\mathrm{ions}} \le \frac{L}{h\,\nu}

The number of ionized molecules is at most the absorbed energy divided by h times the frequency.

The absorbed energy divided by one quantum's energy is the number of quanta absorbed; if each ionizes at most one molecule, that number bounds the ions.

A quantum must at least pay for one ionization

Modern teaching equation, not the equation as printed.

h\,\nu \ge I

h times the frequency is at least the energy to ionize one molecule.

If one quantum ionizes one molecule, only light whose quantum is at least the ionization energy can ionize the gas at all.

Assumptions

One quantum supplies the ionization energy in the stated idealization.
Only the absorbed energy counts, not all the energy that arrives.
A count equality requires one counted ionization per absorbed quantum.

Limits

A threshold is a necessary energy condition, not a material cross-section or a guaranteed event.
An unknown absorbed fraction or yield is never silently taken to be 100 percent.
The energy budget does not predict recombination, collisions, or avalanche multiplication.
The footnote gives no source and no reason for the fivefold ionizing voltage inside the gas, and the passage supplies neither.

Foundations included in this file

Fractions and ratios

What does dividing one quantity by another tell you?

The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6.

When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group.

Worked example: Fourteen metres in ten seconds

A steady walk covers 14 metres in 10 seconds.
Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.

This lesson builds on

Adding and averaging

Return to the chapter

Powers of ten and physical units

How do you write a very small length without counting zeros?

The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm).

The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included.

Worked example: Squaring six micrometres

1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m².

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

Fractions and ratios

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Static laboratory examples

lq-09: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open lq-09 online

Build-time scalar readouts
Quantity	Value	Unit	Status
incidentPower	0.000001	W	value
frequency	2901590000000000	Hz	value
ionizationEnergyPerMolecule	10	eV	value
absorptionEfficiency	0.5	1	value
duration	1	s	value
absorbedLightEnergy	5e-7	J	value
quantumEnergy	1.92261388865385e-18	J	value
quantumEnergyEv	12.000012032717299	eV	value
thresholdFrequency	2417989242084918	Hz	value
thresholdWavelengthNm	123.98419843320026	nm	value
excessEnergyEv	2.0000120327172994	eV	value
singleQuantumAllowed	1	1	value
quantumRate	520125234661.7378	s^-1	value
absorbedQuantumRate	260062617330.8689	s^-1	value
ionizationRate	260062617330.8689	s^-1	value
ionizationCount	260062617330.8689	1	value
ionizedGramMolecules	4.318441359893901e-13	mol	value

0 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

absorptionEfficiency: 0.5
absorptionMode: all-absorbed-ionizes
declaredFraction: 1
duration: 1
frequency: 2901590000000000
gasCitation: P. Lenard, Ann. d. Phys. 3, p. 298, 1900
gasName: Air (Lenard 1900 reference)
incidentPower: 0.000001
ionizationEnergyEv: 10

Evaluator: source:sha256:034358c16972b91b4079c209502bceaca0d9e6fbbda4857396282ee618c308c9

Source references

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

Light quanta: from entropy to an energy scale: §9 · ionization bounds and closing scope