Calculation of wavelength of electron in ground state:




Radius of ground state of hydrogen atom = $0.53\mathit{Å}=0.53\times {10}^{-10}\mathrm{m}$
According to de Broglie relation $2\mathit{\pi }\mathit{r}=\mathit{n}\mathit{\lambda }$
For ground state $\mathit{n}=1$
$2\times 3.14\times 0.53\times {10}^{-10}=1\times \mathit{\lambda }$
$\therefore \mathit{\lambda }=3.32\times {10}^{-10}\mathrm{m}$
$=3.32\mathit{Å}$
Alternatively
Velocity of electron, in the ground state, of hydrogen atom
$=2.18\times {10}^{-6}\mathrm{m}∕\mathrm{}\mathit{s}$
Hence momentum of revolving electron
$\mathit{p}=\mathit{m}\mathit{v}$
$=9.1\times {10}^{-31}\times 2.18$ $\times {10}^{-6}\mathrm{kg}\mathrm{m}∕\mathrm{}\mathit{s}$
$\mathit{\lambda }=\frac{\mathit{h}}{\mathit{p}}$ $=\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}\times 2.18\times {10}^6}\mathrm{m}$
$=3.32\mathit{Å}$
[Note : Also accept the following answer:
Let ${\mathit{\lambda }}_{\mathit{n}}$ be the wavelength of the electron in the orbit, we then have
$2\mathit{\pi }{\mathit{r}}_{\mathit{n}}=\mathit{n}\mathit{\lambda }$
For ground state $\mathit{n}=1$
$2\mathit{\pi }{\mathit{r}}_{\mathit{n}}=\mathit{\lambda }$
( $\mathit{r}={\mathit{r}}_0$ is the radius of the ground state)
[Alternatively
${\mathit{\lambda }}_{\mathit{n}}=\frac{\mathit{h}}{\mathit{m}{\mathit{v}}_{\mathit{n}}}$
and ${\mathit{v}}_{\mathit{n}}={\mathit{v}}_0$ (velocity of electron in ground state)
$\mathit{\lambda }=\frac{\mathit{h}}{\mathit{m}{\mathit{v}}_{\mathit{n}}}$