Question
According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?

Answer

Frequency of the revolution in the ground state is $\frac{\text{V}_0}{2\pi\text{r}_0}$
[$r_0=$ radius of ground state, $V_0=$ velocity in the ground state]
$\therefore$ Frequency of radiation emitted is $\frac{\text{V}_0}{2\pi\text{r}_0}=\text{f}$
$\therefore\text{C}=\text{f}\lambda$
$\lambda=\frac{\text{C}}{\text{f}}=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\therefore\lambda=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\lambda=45.686\text{nm}=45.7\text{nm}$

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