Physics - Modern Physics Question with Solution | TestHub

According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by

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

The de Broglie wavelength of the electron is

$\lambda =\frac{h}{p}=\frac{h}{mv}$

Substituting the value of $mv$ from the first equation into the second equation, we get

$\lambda =\frac{h}{\left(\frac{nh}{2\pi r}\right)}$

$\lambda =\frac{2\pi r}{n}$

Answer: $\frac{2\pi r}{n}$