Based on Maxwell's theory of electrodynamics, an electron...

The frequency of revolution of an electron in its orbit is given by $f = \dfrac{v}{2\pi r}$.

According to Maxwell's theory of electrodynamics, the frequency of the emitted radiation is equal to the frequency of revolution of the electron.

The wavelength of the emitted radiation is $\lambda = \dfrac{c}{f} = \dfrac{2\pi r c}{v}$.

For a hydrogen atom in its ground state ($n=1$):
Radius of the first orbit, $r = 0.529 \times 10^{-10} \text{ m}$
Velocity of the electron, $v = 2.18 \times 10^6 \text{ m/s}$
Speed of light, $c = 3 \times 10^8 \text{ m/s}$

Substituting these values into the expression for wavelength:
$\lambda = \dfrac{2 \times 3.14 \times 0.529 \times 10^{-10} \times 3 \times 10^8}{2.18 \times 10^6}$

$\lambda \approx 45.7 \times 10^{-8} \text{ m}$

$\lambda = 45.7 \text{ nm}$

Answer: $45.7 \text{ nm}$