Page:Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt.pdf/15

 will have lost some of its energy when it arrives at the surface. Other than that we must assume that on leaving the solid every electron must do an amount of work P (characteristic of that solid). Electrons residing right at the surface, excited at right angles to it, will leave the solid with the largest normal velocity. The kinetic energy of such electrons is


 * $$\frac{R}{N}\beta\nu - P.$$

If the body is charged to a positive potential &Pi; and surrounded by conductors with potential zero and &Pi; is just enough to prevent loss of electricity by the body, then we must have:


 * $$\Pi \epsilon = \frac{R}{N}\beta\nu - P ,$$

where &epsilon; is the electrical mass of the electron, or


 * $$\Pi E = R \beta \nu - P^\prime \,\! ,$$

where E is the charge of one gram equivalent of a single-valued ion and P' is the potential of this amount of negative electricity with respect to this body.

If we set E = 9.6&middot;103, then &Pi;&middot;10-8 is the potential in volts that the body will attain when it is irradiated in vacuum.

To see now whether the derived relation agrees with experiment to within an order of magnitude we set P' = 0, &nu; = 1.03&middot;1015 (corresponding to the ultraviolet limit of the solar spectrum), and &beta; = 4.866&middot;10-11. We obtain &Pi;&middot;107 = 4.3 Volt, which agrees to within an order of magnitude with the results of Mr. Lenard.

If the formula derived is correct, then &Pi;, as a function of frequency of the excited light represented in Cartesian coordinates, must be a straight line, whose inclination is independent from the nature of the substance investigated.