On a Heuristic Point of View Concerning the Generation and Transformation of Light

Between the theoretical conceptions that physicists have formed of gases and other ponderable bodies and Maxwell’s theory of electromagnetic processes in so-called empty space there is a profound formal difference.Show the German source For while we regard the state of a body as completely determined by the positions and velocities of a very large, but nevertheless finite, number of atoms and electrons, we make use of continuous spatial functions to determine the electromagnetic state of a space, so that a finite number of quantities is thus not to be regarded as sufficient for the complete specification of the electromagnetic state of a space.Show the German source According to Maxwell’s theory, in all purely electromagnetic phenomena, and so also in light, the energy is to be conceived as a continuous function of space, whereas the energy of a ponderable body, according to the present conception of physicists, is to be represented as a sum extended over the atoms and electrons.Show the German source The energy of a ponderable body cannot break up into arbitrarily many, arbitrarily small parts, whereas the energy of a ray of light emitted from a point source of light, according to Maxwell’s theory (or, more generally, according to any undulatory theory) of light, distributes itself continuously over an ever-increasing volume.Show the German source

The undulatory theory of light, which operates with continuous functions of space, has proved itself excellently in the representation of purely optical phenomena and will probably never be replaced by another theory.Show the German source It should be kept in mind, however, that optical observations refer to time averages and not to instantaneous values, and it is quite conceivable, despite the complete confirmation of the theory of diffraction, reflection, refraction, dispersion, etc. by experiment, that the theory of light that operates with continuous functions of space leads to contradictions with experience when it is applied to the phenomena of the generation and transformation of light.Show the German source

It seems to me, in fact, that the observations of “black radiation”, photoluminescence, the production of cathode rays by ultraviolet light and other groups of phenomena concerning the generation or transformation of light appear more readily understandable on the assumption that the energy of light is distributed discontinuously in space.Show the German source According to the assumption to be considered here, in the propagation of a ray of light issuing from a point the energy is not distributed continuously over ever larger spaces, but consists of a finite number of energy quanta localized at points in space, which move without dividing and can be absorbed and generated only as wholes.Show the German source

In what follows I shall communicate the train of thought and cite the facts that led me onto this path, in the hope that the point of view to be set forth may prove useful to some investigators in their researches.Show the German source

§ 1. On a difficulty concerning the theory of “black radiation”.Show the German source

We first take up the standpoint of Maxwell’s theory and the electron theory and consider the following case.Show the German source In a space enclosed by perfectly reflecting walls, let there be a number of gas molecules and electrons that are free to move and exert conservative forces on one another when they come very close to one another, i.e., that can collide with one another like gas molecules according to the kinetic theory of gases.^[1)]Show the German source Further, let a number of electrons be bound to points of the space far apart from one another, by forces directed toward these points and proportional to the elongations.Show the German source These electrons, too, are to enter into conservative interaction with the free molecules and electrons when the latter come very close to them.Show the German source We call the electrons bound to points in space “resonators”; they emit and absorb electromagnetic waves of definite period.Show the German source

According to the present view of the origin of light, the radiation in the space under consideration, which is found on the basis of Maxwell’s theory for the case of dynamic equilibrium, would have to be identical with the “black radiation” — at least if resonators of all frequencies that come into consideration are regarded as present.Show the German source

For the time being we disregard the radiation emitted and absorbed by the resonators and ask for the condition for dynamic equilibrium that corresponds to the interaction (the collisions) of molecules and electrons.Show the German source For the latter the kinetic theory of gases yields the condition that the mean vis viva of a resonator electron must be equal to the mean kinetic energy of the translational motion of a gas molecule.Show the German source If we resolve the motion of the resonator electron into three mutually perpendicular oscillatory motions, we find for the mean value $\overline{E}$ of the energy of one such rectilinear oscillatory motionShow the German source $$\overline{E} = \frac{R}{N} T,$$$\overline{E}$Mean resonator energy$R$Gas constant$N$Avogadro's number$T$Temperature of the radiationExplain this equationShow the German source where $R$ denotes the absolute gas constant, $N$ the number of “real molecules” in one gram-equivalent and $T$ the absolute temperature.Show the German source For, because the time averages of the kinetic and potential energy of the resonator are equal, the energy $\overline{E}$ is $2/3$ times as large as the vis viva of a free, monatomic gas molecule.Show the German source Now if some cause — in our case, radiation processes — were to bring it about that the energy of a resonator had a greater or smaller time average than $\overline{E}$, then the collisions with the free electrons and molecules would lead to a release of energy to the gas, or an uptake of energy from the gas, that differs from zero on average.Show the German source Thus, in the case we are considering, dynamic equilibrium is possible only if every resonator has the mean energy $\overline{E}$.Show the German source

We now make a similar consideration regarding the interaction of the resonators and the radiation present in the space.Show the German source Mr. Planck has derived the condition of dynamic equilibrium for this case^[1)] on the supposition that the radiation can be regarded as the most disordered process conceivable.^[2)]Show the German source He found:Show the German source $$\overline{E}_\nu = \frac{L^3}{8 \pi \nu^2} \varrho_\nu.$$${\overline{E}}_{\nu }$Mean resonator energy at a given frequency$L$Speed of light$\nu$Frequency of the light${\varrho }_{\nu }$Spectral energy densityExplain this equationShow the German source Here $\overline{E}_\nu$ is the mean energy of a resonator of natural frequency $\nu$ (per component of oscillation), $L$ the velocity of light, $\nu$ the frequency and $\varrho_\nu d \nu$ the energy per unit volume of that part of the radiation whose oscillation number lies between $\nu$ and $\nu + d \nu$.Show the German source

If the radiation energy of frequency $\nu$ is not to be continually either diminished or increased as a whole, then the following must hold:Show the German source $$\frac{R}{N} T = \overline{E} = \overline{E}_\nu = \frac{L^3}{8 \pi \nu^2} \varrho_\nu,$$$R$Gas constant$N$Avogadro's number$T$Temperature of the radiation$\overline{E}$Mean resonator energy${\overline{E}}_{\nu }$Mean resonator energy at a given frequency$L$Speed of light$\nu$Frequency of the light${\varrho }_{\nu }$Spectral energy densityExplain this equationShow the German source $$\varrho_\nu = \frac{R}{N} \frac{8 \pi \nu^2}{L^3} T.$$${\varrho }_{\nu }$Spectral energy density$R$Gas constant$N$Avogadro's number$\nu$Frequency of the light$L$Speed of light$T$Temperature of the radiationExplain this equationShow the German source This relation, found as the condition of dynamic equilibrium, not only lacks agreement with experience, but it also says that in our picture there can be no question of a definite distribution of energy between ether and matter.Show the German source For the wider the range of oscillation numbers of the resonators is chosen, the greater the radiation energy of the space becomes, and in the limit we obtainShow the German source $$\int_0^\infty \varrho_\nu d \nu = \frac{R}{N} \frac{8 \pi}{L^3} T \int_0^\infty \nu^2 d \nu = \infty.$$${\varrho }_{\nu }$Spectral energy density$\nu$Frequency of the light$R$Gas constant$N$Avogadro's number$L$Speed of light$T$Temperature of the radiationExplain this equationShow the German source

§ 2. On Planck’s determination of the elementary quanta.Show the German source

We shall show in what follows that the determination of the elementary quanta given by Mr. Planck is, to a certain extent, independent of the theory of “black radiation” he set up.Show the German source

Planck’s formula^[1)] for $\varrho_\nu$, which satisfies all experience so far, readsShow the German source $$\varrho_\nu = \frac{\alpha \nu^3}{e^{\frac{\beta \nu}{T}} - 1},$$${\varrho }_{\nu }$Spectral energy density$\alpha$Spectral constant A$\nu$Frequency of the light$\beta$Spectral constant B$T$Temperature of the radiationExplain this equationShow the German source whereShow the German source $$\alpha = 6{,}10 \cdot 10^{-56},$$$\alpha$Spectral constant AExplain this equationShow the German source $$\beta = 4{,}866 \cdot 10^{-11}.$$$\beta$Spectral constant BExplain this equationShow the German source For large values of $T/\nu$, i.e., for large wavelengths and radiation densities, this formula goes over in the limit into the following:Show the German source $$\varrho_\nu = \frac{\alpha}{\beta} \nu^2 T.$$${\varrho }_{\nu }$Spectral energy density$\alpha$Spectral constant A$\beta$Spectral constant B$\nu$Frequency of the light$T$Temperature of the radiationExplain this equationShow the German source One sees that this formula agrees with the one developed in § 1 from Maxwell’s theory and the electron theory.Show the German source By equating the coefficients of the two formulas one obtains:Show the German source $$\frac{R}{N} \frac{8 \pi}{L^3} = \frac{\alpha}{\beta}$$$R$Gas constant$N$Avogadro's number$L$Speed of light$\alpha$Spectral constant A$\beta$Spectral constant BExplain this equationShow the German source orShow the German source $$N = \frac{\beta}{\alpha} \frac{8 \pi R}{L^3} = 6{,}17 \cdot 10^{23},$$$N$Avogadro's number$\beta$Spectral constant B$\alpha$Spectral constant A$R$Gas constant$L$Speed of lightExplain this equationShow the German source i.e., an atom of hydrogen weighs $1/N$ grams $= 1{,}62 \cdot 10^{-24}$ g.Show the German source This is exactly the value found by Mr. Planck, which agrees satisfactorily with the values for this quantity found by other means.Show the German source

We therefore arrive at the conclusion: the greater the energy density and the wavelength of a radiation, the more useful do the theoretical foundations we have used prove to be; for small wavelengths and small radiation densities, however, they fail completely.Show the German source

In what follows, “black radiation” will be considered in connection with experience, without taking as a basis any picture of the generation and propagation of the radiation.Show the German source

§ 3. On the entropy of radiation.Show the German source

The following consideration is contained in a famous paper by Mr. W. Wien and is given a place here only for the sake of completeness.Show the German source

Let there be given a radiation that occupies the volume $v$.Show the German source We assume that the perceptible properties of the radiation in question are completely determined when the radiation density $\varrho(\nu)$ is given for all frequencies.^[1)]Show the German source Since radiations of different frequencies are to be regarded as separable from one another without the performance of work and without the supply of heat, the entropy of the radiation can be represented in the formShow the German source $$S = v \int_0^\infty \varphi(\varrho, \nu) d \nu$$$S$Radiation entropy$v$Volume$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source where $\varphi$ denotes a function of the variables $\varrho$ and $\nu$.Show the German source $\varphi$ can be reduced to a function of only one variable by formulating the statement that the entropy of a radiation is not changed by its adiabatic compression between reflecting walls.Show the German source We shall not go into this, however, but shall at once investigate how the function $\varphi$ can be determined from the radiation law of the black body.Show the German source

In “black radiation”, $\varrho$ is such a function of $\nu$ that the entropy is a maximum for a given energy, i.e., thatShow the German source $$\delta \int_0^\infty \varphi(\varrho, \nu) d \nu = 0,$$$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source ifShow the German source $$\delta \int_0^\infty \varrho d \nu = 0.$$$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source From this it follows that for every choice of $\delta \varrho$ as a function of $\nu$Show the German source $$\int_0^\infty \left( \frac{\partial \varphi}{\partial \varrho} - \lambda \right) \delta \varrho d \nu = 0,$$$\phi$Spectral entropy density$\varrho$Spectral energy density$\lambda$Lagrange multiplier$\nu$Frequency of the lightExplain this equationShow the German source where $\lambda$ is independent of $\nu$.Show the German source In black radiation, therefore, $\partial \varphi / \partial \varrho$ is independent of $\nu$.Show the German source

For the increase in temperature of a black radiation of volume $v = 1$ by $d T$, the following equation holds:Show the German source $$d S = \int_{\nu = 0}^{\nu = \infty} \frac{\partial \varphi}{\partial \varrho} d \varrho d \nu,$$$S$Radiation entropy$\nu$Frequency of the light$\phi$Spectral entropy density$\varrho$Spectral energy densityExplain this equationShow the German source or, since $\partial \varphi / \partial \varrho$ is independent of $\nu$:Show the German source $$d S = \frac{\partial \varphi}{\partial \varrho} d E.$$$S$Radiation entropy$\phi$Spectral entropy density$\varrho$Spectral energy density$E$Radiation energy in the bandExplain this equationShow the German source Since $d E$ is equal to the heat supplied and the process is reversible, we also have:Show the German source $$d S = \frac{1}{T} d E.$$$S$Radiation entropy$T$Temperature of the radiation$E$Radiation energy in the bandExplain this equationShow the German source By comparison one obtains:Show the German source $$\frac{\partial \varphi}{\partial \varrho} = \frac{1}{T}.$$$\phi$Spectral entropy density$\varrho$Spectral energy density$T$Temperature of the radiationExplain this equationShow the German source This is the law of black radiation.Show the German source One can therefore determine the law of black radiation from the function $\varphi$ and, conversely, the function $\varphi$ from the latter by integration, bearing in mind that $\varphi$ vanishes for $\varrho = 0$.Show the German source

§ 4. Limiting law for the entropy of monochromatic radiation at low radiation density.Show the German source

It does follow, to be sure, from the observations of “black radiation” so far that the law originally set up by Mr. W. Wien for “black radiation”Show the German source $$\varrho = \alpha \nu^3 e^{-\beta \frac{\nu}{T}}$$$\varrho$Spectral energy density$\alpha$Spectral constant A$\nu$Frequency of the light$\beta$Spectral constant B$T$Temperature of the radiationExplain this equationShow the German source is not exactly valid.Show the German source But for large values of $\nu/T$ it has been confirmed very completely by experiment.Show the German source We take this formula as the basis of our calculations, but bear in mind that our results hold only within certain limits.Show the German source

From this formula there follows first:Show the German source $$\frac{1}{T} = -\frac{1}{\beta \nu} \lg \frac{\varrho}{\alpha \nu^3}$$$T$Temperature of the radiation$\beta$Spectral constant B$\nu$Frequency of the light$\varrho$Spectral energy density$\alpha$Spectral constant AExplain this equationShow the German source and further, using the relation found in the preceding section:Show the German source $$\varphi(\varrho, \nu) = -\frac{\varrho}{\beta \nu} \left\{ \lg \frac{\varrho}{\alpha \nu^3} - 1 \right\}.$$$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the light$\beta$Spectral constant B$\alpha$Spectral constant AExplain this equationShow the German source Now let there be given a radiation of energy $E$ whose frequency lies between $\nu$ and $\nu + d \nu$.Show the German source Let the radiation occupy the volume $v$.Show the German source The entropy of this radiation is:Show the German source $$S = v \varphi(\varrho, \nu) d \nu = -\frac{E}{\beta \nu} \left\{ \lg \frac{E}{v \alpha \nu^3 d \nu} - 1 \right\}.$$$S$Radiation entropy$v$Volume$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the light$E$Radiation energy in the band$\beta$Spectral constant B$\alpha$Spectral constant AExplain this equationShow the German source If we confine ourselves to investigating the dependence of the entropy on the volume occupied by the radiation, and denote by $S_0$ the entropy of the radiation when it has the volume $v_0$, we obtain:Show the German source $$S - S_0 = \frac{E}{\beta \nu} \lg \left( \frac{v}{v_0} \right).$$$S$Radiation entropy$S_0$Radiation entropy$E$Radiation energy in the band$\beta$Spectral constant B$\nu$Frequency of the light$v$Volume$v_0$VolumeExplain this equationShow the German source

This equation shows that the entropy of a monochromatic radiation of sufficiently small density varies with the volume according to the same law as the entropy of an ideal gas or that of a dilute solution.Show the German source The equation just found will be interpreted in what follows on the basis of the principle introduced into physics by Mr. Boltzmann, according to which the entropy of a system is a function of the probability of its state.Show the German source

§ 5. Molecular-theoretical investigation of the dependence of the entropy of gases and dilute solutions on the volume.Show the German source

In calculating entropy by molecular-theoretical methods, the word “probability” is often used in a sense that does not coincide with the definition of probability as it is given in the calculus of probabilities.Show the German source In particular, the “cases of equal probability” are often fixed hypothetically in cases where the theoretical pictures used are definite enough to give a deduction in place of that hypothetical stipulation.Show the German source I shall show in a separate paper that in considerations of thermal processes one manages completely with the so-called “statistical probability”, and I hope thereby to remove a logical difficulty that still stands in the way of carrying through Boltzmann’s principle.Show the German source Here, however, only its general formulation and its application to quite special cases will be given.Show the German source

If it makes sense to speak of the probability of a state of a system, and if, further, every increase in entropy can be conceived as a transition to a more probable state, then the entropy $S_1$ of a system is a function of the probability $W_1$ of its momentary state.Show the German source If, therefore, two systems $S_1$ and $S_2$ are given that do not interact with each other, one can set:Show the German source $$\begin{aligned} S_1 &= \varphi_1(W_1), \\ S_2 &= \varphi_2(W_2). \end{aligned}$$$S_1$Entropy${\phi }_1$Entropy$W_1$Probability of the configuration$S_2$Entropy${\phi }_2$Entropy$W_2$Probability of the configurationExplain this equationShow the German source If one regards these two systems as a single system of entropy $S$ and probability $W$, then we have:Show the German source $$S = S_1 + S_2 = \varphi(W)$$$S$Entropy$S_1$Entropy$S_2$Entropy$\phi$Entropy$W$Probability of the configurationExplain this equationShow the German source andShow the German source $$W = W_1 \cdot W_2.$$$W$Probability of the configuration$W_1$Probability of the configuration$W_2$Probability of the configurationExplain this equationShow the German source The latter relation says that the states of the two systems are events independent of each other.Show the German source

From these equations it follows:Show the German source $$\varphi(W_1 \cdot W_2) = \varphi_1(W_1) + \varphi_2(W_2)$$$\phi$Entropy$W_1$Probability of the configuration$W_2$Probability of the configuration${\phi }_1$Entropy${\phi }_2$EntropyExplain this equationShow the German source and from this, finally,Show the German source $$\begin{aligned} \varphi_1(W_1) &= C \lg(W_1) + \text{konst.}, \\ \varphi_2(W_2) &= C \lg(W_2) + \text{konst.}, \\ \varphi(W) &= C \lg(W) + \text{konst.} \end{aligned}$$${\phi }_1$Entropy$W_1$Probability of the configuration$C$Universal entropy constant${\phi }_2$Entropy$W_2$Probability of the configuration$\phi$Entropy$W$Probability of the configurationExplain this equationShow the German source The quantity $C$ is therefore a universal constant; as follows from the kinetic theory of gases, it has the value $R/N$, where the constants $R$ and $N$ are to be given the same meaning as above.Show the German source If $S_0$ denotes the entropy in a certain initial state of a system under consideration and $W$ the relative probability of a state of entropy $S$, we thus obtain in general:Show the German source $$S - S_0 = \frac{R}{N} \lg W.$$$S$Entropy$S_0$Entropy$R$Gas constant$N$Avogadro's number$W$Probability of the configurationExplain this equationShow the German source

We first treat the following special case.Show the German source In a volume $v_0$ let there be a number $(n)$ of movable points (e.g., molecules) to which our consideration is to refer.Show the German source Besides these, arbitrarily many other movable points of any kind may be present in the space.Show the German source About the law by which the points under consideration move in the space, nothing is presupposed except that, with respect to this motion, no part of the space (and no direction) is distinguished from the others.Show the German source Further, let the number of the (first-mentioned) movable points under consideration be so small that any action of the points on one another can be disregarded.Show the German source

The system under consideration, which may be, e.g., an ideal gas or a dilute solution, has a certain entropy $S_0$.Show the German source We imagine a part of the volume $v_0$ of size $v$, and all $n$ movable points transferred into the volume $v$, without anything else in the system being changed.Show the German source To this state there evidently belongs a different value of the entropy $(S)$, and we now want to determine the difference in entropy with the help of Boltzmann’s principle.Show the German source

We ask: How large is the probability of the last-mentioned state relative to the original one?Show the German source Or: How large is the probability that at a randomly chosen instant of time all $n$ points movable independently of one another in a given volume $v_0$ are (by chance) in the volume $v$?Show the German source

For this probability, which is a “statistical probability”, one evidently obtains the value:Show the German source $$W = \left( \frac{v}{v_0} \right)^n ;$$$W$Probability of the configuration$v$Volume$v_0$Volume$n$Number of independent pointsExplain this equationShow the German source from this one obtains, by applying Boltzmann’s principle:Show the German source $$S - S_0 = R \left( \frac{n}{N} \right) \lg \left( \frac{v}{v_0} \right).$$$S$Entropy$S_0$Entropy$R$Gas constant$n$Number of independent points$N$Avogadro's number$v$Volume$v_0$VolumeExplain this equationShow the German source

It is remarkable that in deriving this equation, from which the Boyle-Gay-Lussac law and the law of the same form for osmotic pressure can easily be derived thermodynamically^[1)]; one need make no assumption about the law according to which the molecules move.Show the German source

§ 6. Interpretation of the expression for the dependence of the entropy of monochromatic radiation on the volume according to Boltzmann’s principle.Show the German source

In § 4 we found for the dependence of the entropy of monochromatic radiation on the volume the expression:Show the German source $$S - S_0 = \frac{E}{\beta \nu} \lg \left( \frac{v}{v_0} \right).$$$S$Radiation entropy$S_0$Radiation entropy$E$Radiation energy in the band$\beta$Spectral constant B$\nu$Frequency of the light$v$Volume$v_0$VolumeExplain this equationShow the German source If one writes this formula in the form:Show the German source $$S - S_0 = \frac{R}{N} \lg \left[ \left( \frac{v}{v_0} \right)^{\frac{N}{R} \frac{E}{\beta \nu}} \right]$$$S$Radiation entropy$S_0$Radiation entropy$R$Gas constant$N$Avogadro's number$v$Volume$v_0$Volume$E$Radiation energy in the band$\beta$Spectral constant B$\nu$Frequency of the lightExplain this equationShow the German source and compares it with the general formula expressing Boltzmann’s principleShow the German source $$S - S_0 = \frac{R}{N} \lg W ,$$$S$Radiation entropy$S_0$Radiation entropy$R$Gas constant$N$Avogadro's number$W$Probability of the configurationExplain this equationShow the German source one arrives at the following conclusion:Show the German source

If monochromatic radiation of frequency $\nu$ and energy $E$ is enclosed in the volume $v_0$ (by reflecting walls), the probability that at an arbitrarily chosen instant of time the whole radiation energy is in the partial volume $v$ of the volume $v_0$ is:Show the German source $$W = \left( \frac{v}{v_0} \right)^{\frac{N}{R} \frac{E}{\beta \nu}}.$$$W$Probability of the configuration$v$Volume$v_0$Volume$N$Avogadro's number$R$Gas constant$E$Radiation energy in the band$\beta$Spectral constant B$\nu$Frequency of the lightExplain this equationShow the German source

From this we further conclude:Show the German source

Monochromatic radiation of low density (within the range of validity of Wien’s radiation formula) behaves in thermal respects as if it consisted of mutually independent energy quanta of magnitude $R \beta \nu / N$.Show the German source

We shall also compare the mean magnitude of the energy quanta of “black radiation” with the mean vis viva of the center-of-mass motion of a molecule at the same temperature.Show the German source The latter is $\frac{3}{2} (R/N) T$, while for the mean magnitude of the energy quantum one obtains, on the basis of Wien’s formula:Show the German source $$\frac{\int_0^\infty \alpha \nu^3 e^{-\frac{\beta \nu}{T}} d \nu}{\int_0^\infty \frac{N}{R \beta \nu} \alpha \nu^3 e^{-\frac{\beta \nu}{T}} d \nu} = 3 \frac{R}{N} T.$$$\alpha$Spectral constant A$\nu$Frequency of the light$\beta$Spectral constant B$T$Temperature of the radiation$N$Avogadro's number$R$Gas constantExplain this equationShow the German source

If, now, monochromatic radiation (of sufficiently small density) behaves, as regards the dependence of the entropy on the volume, like a discontinuous medium consisting of energy quanta of magnitude $R \beta \nu / N$, it is natural to investigate whether the laws of the generation and transformation of light are also such as they would be if light consisted of energy quanta of this kind.Show the German source We shall occupy ourselves with this question in what follows.Show the German source

§ 7. On Stokes’s rule.Show the German source

Let monochromatic light be transformed by photoluminescence into light of a different frequency, and, in accordance with the result just obtained, let it be assumed that both the generating and the generated light consist of energy quanta of magnitude $(R/N) \beta \nu$, where $\nu$ denotes the frequency in question.Show the German source The transformation process is then to be interpreted as follows.Show the German source Each generating energy quantum of frequency $\nu_1$ is absorbed and — at least when the distribution density of the generating energy quanta is sufficiently small — by itself alone gives rise to the production of a light quantum of frequency $\nu_2$; possibly, in the absorption of the generating light quantum, light quanta of frequencies $\nu_3$, $\nu_4$, etc., as well as energy of other kinds (e.g., heat) may also arise at the same time.Show the German source Through what intermediate processes this final result comes about is immaterial.Show the German source If the photoluminescent substance is not to be regarded as a permanent source of energy, then by the energy principle the energy of a generated energy quantum cannot be greater than that of a generating light quantum; the following relation must therefore hold:Show the German source $$\frac{R}{N} \beta \nu_2 \leqq \frac{R}{N} \beta \nu_1$$$R$Gas constant$N$Avogadro's number$\beta$Spectral constant B${\nu }_2$Frequency of the emitted light${\nu }_1$Frequency of the absorbed lightExplain this equationShow the German source orShow the German source $$\nu_2 \leqq \nu_1 .$$${\nu }_2$Frequency of the emitted light${\nu }_1$Frequency of the absorbed lightExplain this equationShow the German source This is Stokes’s well-known rule.Show the German source

It is especially to be emphasized that with weak illumination, other circumstances being equal, the quantity of light generated must according to our conception be proportional to the intensity of the exciting light, since each exciting energy quantum will cause an elementary process of the kind indicated above, independently of the action of the other exciting energy quanta.Show the German source In particular, there will be no lower limit for the intensity of the exciting light below which the light would be incapable of exciting light.Show the German source

According to the conception of the phenomena set forth here, deviations from Stokes’s rule are conceivable in the following cases:
1. when the number of energy quanta per unit volume simultaneously undergoing transformation is so large that an energy quantum of the generated light can receive its energy from several generating energy quanta;
2. when the generating (or generated) light does not have the energetic character belonging to a “black radiation” within the range of validity of Wien’s law, that is, when, e.g., the exciting light is produced by a body of so high a temperature that Wien’s law no longer holds for the wavelength in question.Show the German source

The last-mentioned possibility deserves particular interest.Show the German source For according to the conception developed here it is not ruled out that a “non-Wien radiation”, even at great dilution, behaves energetically differently from a “black radiation” within the range of validity of Wien’s law.Show the German source

§ 8. On the production of cathode rays by the illumination of solid bodies.Show the German source

The usual conception, that the energy of light is distributed continuously over the space through which it passes, meets with especially great difficulties in the attempt to explain the photoelectric phenomena, difficulties which have been set out in a path-breaking paper by Mr. Lenard.^[1)]Show the German source

On the conception that the exciting light consists of energy quanta of energy $(R/N) \beta \nu$, the production of cathode rays by light can be conceived as follows.Show the German source Energy quanta penetrate into the surface layer of the body, and their energy is transformed, at least in part, into kinetic energy of electrons.Show the German source The simplest picture is that a light quantum gives up its entire energy to a single electron; we shall assume that this occurs.Show the German source It is not to be excluded, however, that electrons take up the energy of light quanta only in part.Show the German source An electron provided with kinetic energy in the interior of the body will have lost part of its kinetic energy by the time it has reached the surface.Show the German source Furthermore, it is to be assumed that in leaving the body each electron has to perform an amount of work $P$ (characteristic of the body) when it leaves the body.Show the German source The electrons excited directly at the surface and normal to it will leave the body with the greatest normal velocity.Show the German source The kinetic energy of such electrons isShow the German source $$\frac{R}{N} \beta \nu - P .$$$R$Gas constant$N$Avogadro's number$\beta$Spectral constant B$\nu$Frequency of the light$P$Escape workExplain this equationShow the German source

If the body is charged to the positive potential $\Pi$ and surrounded by conductors at zero potential, and if $\Pi$ is just able to prevent a loss of electricity from the body, then we must have:Show the German source $$\Pi \varepsilon = \frac{R}{N} \beta \nu - P ,$$$\Pi$Stopping potential$\epsilon$Charge of one electron$R$Gas constant$N$Avogadro's number$\beta$Spectral constant B$\nu$Frequency of the light$P$Escape workExplain this equationShow the German source where $\varepsilon$ denotes the electric mass of the electron, orShow the German source $$\Pi E = R \beta \nu - P' ,$$$\Pi$Stopping potential$E$Gram-equivalent charge$R$Gas constant$\beta$Spectral constant B$\nu$Frequency of the light$P^{\prime }$Work function per gram-equivalentExplain this equationShow the German source where $E$ denotes the charge of one gram-equivalent of a univalent ion and $P'$ the potential of this quantity of negative electricity with respect to the body.^[1)]Show the German source

If one sets $E = 9{,}6 \cdot 10^3$, then $\Pi \cdot 10^{-8}$ is the potential in volts that the body assumes when irradiated in vacuum.Show the German source

In order to see first whether the relation derived agrees with experience as to order of magnitude, we set $P' = 0$, $\nu = 1{,}03 \cdot 10^{15}$ (corresponding to the limit of the solar spectrum toward the ultraviolet) and $\beta = 4{,}866 \cdot 10^{-11}$.Show the German source We obtain $\Pi \cdot 10^7 = 4{,}3$ volts, a result that agrees in order of magnitude with Mr. Lenard’s results.^[2)]Show the German source

If the formula derived is correct, then $\Pi$, represented in Cartesian coordinates as a function of the frequency of the exciting light, must be a straight line whose slope is independent of the nature of the substance investigated.Show the German source

So far as I can see, our conception does not stand in contradiction to the properties of the photoelectric effect observed by Mr. Lenard.Show the German source If each energy quantum of the exciting light gives up its energy to electrons independently of all the others, then the velocity distribution of the electrons, i.e., the quality of the cathode radiation produced, will be independent of the intensity of the exciting light; on the other hand, the number of electrons leaving the body will, other circumstances being equal, be proportional to the intensity of the exciting light.^[1)]Show the German source

About the presumable limits of validity of the regularities mentioned, remarks would have to be made similar to those about the presumable deviations from Stokes’s rule.Show the German source

In the foregoing it has been assumed that the energy of at least some of the energy quanta of the generating light is given up completely, each to a single electron.Show the German source If one does not make this natural assumption, one obtains, instead of the above equation, the following:Show the German source

For cathode luminescence, which constitutes the inverse process to the one just considered, one obtains by a consideration analogous to the one carried out:Show the German source

For the substances investigated by Mr. Lenard, $P E$ is always considerably greater than $R \beta \nu$, since the voltage that the cathode rays must have traversed in order to be able to produce just visible light amounts in some cases to a few hundred, in others to thousands of volts.^[2)]Show the German source It is therefore to be assumed that the kinetic energy of one electron is used for the production of many light energy quanta.Show the German source

§ 9. On the ionization of gases by ultraviolet light.Show the German source

We shall have to assume that in the ionization of a gas by ultraviolet light one absorbed light energy quantum is used for the ionization of one gas molecule.Show the German source From this it follows first that the ionization work (i.e., the work theoretically necessary for ionization) of a molecule cannot be greater than the energy of an absorbed effective light energy quantum.Show the German source If one denotes by $J$ the (theoretical) ionization work per gram-equivalent, then we must have:Show the German source $$R \beta \nu \geqq J .$$$R$Gas constant$\beta$Spectral constant B$\nu$Frequency of the light$J$Ionization work per gram-equivalentExplain this equationShow the German source According to Lenard’s measurements, however, the greatest effective wavelength for air is about $1{,}9 \cdot 10^{-5}$ cm, henceShow the German source $$R \beta \nu = 6{,}4 \cdot 10^{12} \text{ Erg} \geqq J .$$$R$Gas constant$\beta$Spectral constant B$\nu$Frequency of the light$J$Ionization work per gram-equivalentExplain this equationShow the German source

An upper limit for the ionization work is also obtained from the ionization voltages in rarefied gases.Show the German source According to J. Stark^[1)] the smallest ionization voltage measured (at platinum anodes) for air is about 10 volts.^[2)]Show the German source For $J$ there thus results the upper limit $9{,}6 \cdot 10^{12}$, which is nearly equal to the one just found.Show the German source There results yet another consequence, whose testing by experiment seems to me to be of great importance.Show the German source If each absorbed light energy quantum ionizes one molecule, then between the absorbed quantity of light $L$ and the number $j$ of gram-molecules ionized by it the following relation must hold:Show the German source $$j = \frac{L}{R \beta \nu} .$$$j$Ionized gram-molecules$L$Light energy absorbed$R$Gas constant$\beta$Spectral constant B$\nu$Frequency of the lightExplain this equationShow the German source This relation must, if our conception corresponds to reality, hold for every gas that (at the frequency in question) shows no appreciable absorption unaccompanied by ionization.Show the German source

Bern, 17 March 1905.Show the German source

(Received 18 March 1905.)Show the German source