On the Attenuation of Atomic Luminescence

(W. Wien. Über die Messungen der Leuchtdauer der Atome und der Dämpfung der Spektrallinien. Ann. der Phys. 60, p. 697—1919).

The author observed the attenuation of the luminescence of a canal ray emerging from a space at high pressure (from 0.06 to 0.014 mm Hg), where the discharge occurred, through a narrow slit (0.1 to 0.2 mm wide) into a vessel at low pressure (from 0.0015 to 0.0003 mm Hg). The canal ray, having emerged into the space with such a low pressure, propagates freely in it, and the luminous atoms present in it gradually fade, losing energy through radiation. The extinguished atoms are not re-excited, since, owing to the reduced pressure, they do not encounter molecules on their path in collision with which the atoms could again begin to radiate.

The canal ray emerging from the above-mentioned narrow slit was set at the focus of the collimator lens of the spectrograph (in place of the withdrawn slit) and was photographed. For comparison, together with the canal ray, the spectrum of hydrogen was photographed. For this purpose a special slit was illuminated, with the aid of a lens, by the light of a hydrogen tube, and the image of this slit (equal to the slit itself) was projected in the same plane as the canal ray at the focus of the same collimator of the spectrograph; the thickness of the slit was made the same as the thickness of the canal ray, and the length of the slit was exactly equal to the length of the canal ray. In the path of the hydrogen rays a V-shaped cuvette was now placed before the slit, so that its edge was perpendicular to the slit; this cuvette was filled with a dye solution, the absorption of which may be regarded as independent of the wavelength in the given part of the spectrum, and, in order to compensate for the absorption of the solvent, a second cuvette of the same kind, but with its edge upward, filled with pure solvent, was placed against the first cuvette. Thus, on the photogra—

on the photographic plate there were obtained: first, the image of an attenuating canal ray, the weakening of whose light was caused by the weakening of the luminescence of the atoms, and the law of attenuation of the canal ray may be expressed as

\[ e^{-2\alpha t}, \]

where \(2\alpha\) is the attenuation constant and \(t\) the time of luminescence; if \(v\) is the velocity of the canal ray, then \(vt=y\), where \(y\) is the distance traversed, and

\[ e^{-2\alpha t}=e^{-2\alpha \frac{y}{v}}. \]

Secondly, on the photographic plate there were obtained images of hydrogen lines weakened on one side by a wedge. The law of attenuation of each of them will be

\[ e^{-ky\tg\beta}, \]

where \(k\) is the constant of absorption by the dye, \(\beta\) is the angle of the wedge, and \(y\) is the distance from the edge of the wedge to the given point. Since the image of the canal ray and the wedge were identical, \(y\) was the same in both cases. If the blackening and its decrease on the photographic plate are selected to be exactly the same for the canal ray and the line of the hydrogen spectrum, then

\[ e^{-2\alpha t}=e^{-2\alpha \frac{y}{v}}=e^{-ky\tg\beta}, \]

whence \(2\alpha=k\gamma v\), where \(\gamma=\tg\beta\). \(\gamma\) and \(k\) are easy to measure, while \(v\) is measured from the Doppler effect, by the formula

\[ v=\frac{\delta\lambda}{\lambda}\,c, \]

where \(\delta\lambda\) is the Doppler shift, and \(c\) is the velocity of light. Thus the author found

\[ \begin{aligned} &\text{for } H\alpha \text{ the value } 2\alpha=6.20\cdot10^{7}\ \mathrm{sec}^{-1},\\ &\text{for } H\beta \quad\text{''}\quad 2\alpha=5.62\cdot10^{7}\ \mathrm{sec}^{-1},\\ &\text{for } H\gamma \quad\text{''}\quad 2\alpha=6.62\cdot10^{7}\ \mathrm{sec}^{-1}. \end{aligned} \]

From the electron theory, by the formula

\[ 2\alpha=\frac{8\pi^{2}e^{2}}{3mc\lambda^{2}}, \]

one obtains

\[ \begin{aligned} &\text{for } H\alpha \text{ the value } 2\alpha=5.35\cdot10^{7}\ \mathrm{sec}^{-1},\\ &\text{for } H\beta \quad\text{''}\quad 2\alpha=9.77\cdot10^{7}\ \mathrm{sec}^{-1},\\ &\text{for } H\gamma \quad\text{''}\quad 2\alpha=12.7\cdot10^{7}\ \mathrm{sec}^{-1}. \end{aligned} \]

Agreement is obtained only for \(H\alpha\), while for the other lines we already see a strong discrepancy; this is understandable: experiment gives one value of the attenuation for all hydrogen lines, whereas according to the electron theory the attenuation is inversely proportional to the square of the wavelength. For oxygen the author obtained:

\[ 2\alpha=6.55\cdot10^{7}\ \mathrm{sec}^{-1}, \]

i.e., equal to the attenuation of the hydrogen atom.

Further, on the basis of the existing theoretical material on the structure of the atom, mainly on the basis of the recent works of Bohr and Sommerfeld, the author attempts to derive consequences that would make it possible to explain the experimental fact obtained by him—the equality of the attenuation for all hydrogen lines—but, as the author himself notes, without success.

N. Shchodro.

1. \(e\) is a positive charge equal in absolute magnitude to the charge of the electron. ↩