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

§ 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

Bern, 17 March 1905.Show the German source

(Received 18 March 1905.)Show the German source