For an electron moving in the $\mathrm{n}^{\text {th }}$ Bohr orbit the deBroglie wavelength of an electron is
For an electron moving in the $\mathrm{n}^{\text {th }}$ Bohr orbit the deBroglie wavelength of an electron is
$\mathrm{n} \pi \mathrm{r}$
$\frac{\pi \mathrm{r}}{\mathrm{n}}$
$\frac{\mathrm{n} \mathrm{r}}{2\pi}$
$\frac{2\pi \mathrm{r}}{\mathrm{n}}$
Solution
From de Broglie's hypothesis,
$\lambda=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{n}}}=\frac{\mathrm{h}}{\mathrm{mv}}$
and from Bohr's atomic model,
$\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$
Also,
$\mathrm{L}=\mathrm{mvr}_{\mathrm{n}}$
Due to quantization of angular momentum, we can write,
$\begin{aligned}
& \frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}} \\
\therefore \quad & \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}
\end{aligned}$
putting (ii) into (i), we get,
$\therefore \quad \lambda=\frac{\mathrm{h}}{\mathrm{m}(\mathrm{nh} / 2 \pi \mathrm{mr})}=\frac{2 \pi \mathrm{r}}{\mathrm{n}}$