152.

The total energy emitted in the time element $dt={\displaystyle \frac{d\rho }{a}}$ is found from [eqn:(258)] (258) by considering that every emitting oscillator expends all its energy of vibration and is

$&N \eta\, d\rho\, (R_{0} + 2R_{1} + 3R_{2} + \dots) \epsilon\ = &N \eta\, d\rho\, \eta(1 + 2(1 - \eta) + 3(1 - \eta)^{2} + \dots)\ = &N\, d\rho = Na\, dt.$

It is therefore equal to the energy absorbed in the same time by all oscillators [eqn:(250)] (250), as is necessary, since the state is one of statistical equilibrium.

Let us now consider the mean energy $\overline{U\mathrm{\backslash Strut}}$ of an oscillator. It is evidently given by the following relation, which is derived in the same way as [eqn:(260)] (260): $N\overline{U\mathrm{\backslash Strut}}=N{\sum }_0^{\infty }{\int }_0^{ϵ}(nϵ+\rho )R_{n}d\rho .\mathrm{\backslash Label}[eqn](263)\text{upshape (263)}$ From this it follows by means of [eqn:(262)] (262), that $\overline{U\mathrm{\backslash Strut}}=(\frac{1}{\eta }-\frac{1}{2})ϵ=(\frac{1}{\eta }-\frac{1}{2})h\nu .\mathrm{\backslash Label}[eqn](264)\text{upshape (264)}$ Since $\eta <1$, $\overline{U\mathrm{\backslash Strut}}$ lies between ${\displaystyle \frac{h\nu }{2}}$ and $\infty$. Indeed, it is immediately evident that $\overline{U\mathrm{\backslash Strut}}$ can never become less than ${\displaystyle \frac{h\nu }{2}}$ since the energy of every oscillator, however small it may be, will assume the value $ϵ=h\nu$ within a time limit, which can be definitely stated.