Page:Über die scheinbare Masse der Ionen.djvu/1

 H. A. Lorentz (Leiden)

It is known that by observations of cathode rays we were able to derive the ratio $$\tfrac{e}{m}$$, i.e. the ratio between the charge of an ion $$e$$ and its mass $$m$$. The question arises, what is meant by that mass. In any case we must attribute an apparent mass to the ion, as it generates a certain energy in the ether by virtue of its motion. This apparent mass will be denoted by $$m_0$$. It is possible that the ion also possesses a real mass in the ordinary sense of the word; in this case, $$m_0 < m$$. If this is not the case, then $$m_0 = m$$.

So we have the inequality

$$\frac{e}{m_{0}}>\frac{e}{m}{,}$$

when there still is a real mass besides the apparent mass; otherwise

$$\frac{e}{m_{0}}=\frac{e}{m}.$$

So we want to write

$$\frac{e}{m_{0}}\geqq\frac{e}{m}{,}$$

where $$\tfrac{e}{m}=10^{7}$$ is.

Now

$$m_{0}=\frac{8}{3}\pi R\sigma e{,}$$

if we conceive the ion as a sphere, $$R$$ is the radius of this sphere, and $$\sigma$$ means the surface density of the charge.

This formula allows for an interesting conclusion on the radius of the ions. If, namely, we substitute for $$m_0$$ the now specified value into the inequality, we obtain an inequality for the radius. We have

$$4\pi R^{2}\cdot\sigma=e{,}$$

thus

$$m_{0}=\frac{8}{3}\pi R\sigma e=\frac{8}{3}\pi Re\cdot\frac{e}{4\pi R^{2}}=\frac{2e^{2}}{3R}$$

and thus

$$\frac{e}{m_{0}}=\frac{3R}{2e}{,}$$

and

$$\frac{3R}{2e}\geq10^{7}$$

and

$$R>10^{7}\cdot\frac{2}{3}e.$$

The magnitude $$e$$ is unfortunately not known. If we take the charge of an ion in a cathode ray to be as great as in an electrolytic hydrogen, and presuppose the size of a hydrogen molecule, we obtain for $$R$$ a magnitude of order