The de-Broglie wavelength of an electron moving in the $n^{ ext {th }}$ Bohr orbit of radius $r$ is

Question

The de-Broglie wavelength of an electron moving in the $n^{	ext {th }}$ Bohr orbit of radius $r$ is

Aicharya · Accepted Answer

The de Broglie wavelength of an electron is given by $\lambda=\frac{h}{m v}$ $\Rightarrow m v=\frac{h}{\lambda} \quad---(1)$
And for the electron moving in the $n^{\text {th }}$ orbit,
The angular momentum $L=\frac{n h}{2 \pi}$
$\begin{aligned}
& \Rightarrow m v r=\frac{n h}{2 \pi} \\
& \Rightarrow m v=\frac{n h}{2 \pi r} \quad---(2)
\end{aligned}$ Taking the ratio of equation (1) and (2), $\frac{h}{\lambda}=\frac{n h}{2 \pi r}$ $\Rightarrow \lambda=\frac{2 \pi r}{n}$ :

The de-Broglie wavelength of an electron moving in the $n^{\text {th }}$ Bohr orbit of radius $r$ is

Solution