Explain the quantization of angular momentum by considering the electron as a wave in an atom.

(N/A) In his second postulate,Bohr proposed that the angular momentum of an electron orbiting the nucleus is quantized.
Louis de Broglie provided a physical explanation for this in $1923$.
According to de Broglie's hypothesis,material particles exhibit wave-like properties. This wave nature was experimentally verified by Davisson and Germer.
Bohr suggested that an electron in a circular orbit behaves as a matter wave.
Similar to waves on a string,particle waves can form standing waves under resonant conditions.
When a string is plucked,many waves are excited,but only those that form standing waves (with nodes at the ends) persist. This occurs when the total distance traveled by the wave is an integral multiple of its wavelength.
Waves with other wavelengths interfere destructively upon reflection,and their amplitudes drop to zero; thus,an electron cannot exist in such an orbit.
For an electron in the $n^{\text{th}}$ circular orbit of radius $r_n$,the circumference of the orbit must be an integral multiple of the wavelength $\lambda$.
Therefore,$2 \pi r_n = n \lambda$ ... $(1)$,where $n = 1, 2, 3, \dots$
Using de Broglie's relation for wavelength and momentum,$\lambda = \frac{h}{p} = \frac{h}{m v_n}$ ... $(2)$.
Substituting $(2)$ into $(1)$:
$2 \pi r_n = n \left( \frac{h}{m v_n} \right)$
Rearranging the terms gives the quantization condition for angular momentum:
$m v_n r_n = \frac{n h}{2 \pi}$