$m \rightarrow$ The mass of the electron and

$v \rightarrow$ The Velocity in the orbit.

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Bohr's Quantization Condition

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The Quantization Condition in Bohr Theory of Hydrogen Atom:
$L=\frac{nh}{2 \pi}$

For the angular momentum L, the electron moves arbitrarily only in a stationary circular orbit. According to De Broglie's hypothesis, this condition can be easily obtained. For this purpose, there are following assumptions given below:

1.) The motion of the electron in a stationary circular orbit is represented by a standing matter-wave. If the wavelength of the wave is $\lambda$ then the De Broglie relation

$\lambda=\frac{h}{mv} \qquad(1)$

Where

$m \rightarrow$ The mass of the electron and

$v \rightarrow$ The Velocity in the orbit.

2.) The circular orbit contains an integral number of wavelengths, i.e.

$2 \pi r_{n}= n \lambda $

$\frac{2 \pi r_{n}}{\lambda}= n \qquad(2)$

Where $n=1,2,3............$ and $r_{n}$ is the radius of the orbit.

Substituting the value of $\lambda$ in equation$(2)$

$\frac{2 \pi r_{n} m v}{h} =n$

$mvr_{n} =\frac{nh}{2\pi}$

$L=\frac{nh}{2\pi}$

Which is Bohr's quantization condition.

$m \rightarrow$ The mass of the electron and

$v \rightarrow$ The Velocity in the orbit.

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