The energy of the state of Hydrogen atom
$\mathrm{En}=\frac{-13.6}{{\mathit{n}}^2}\mathrm{eV}$
For ground state, $\mathit{n}=1$
When atom is in first excited state, $\mathit{n}=2$
$\therefore \mathrm{E}=\frac{-13.6}{2^2}=-3.4\mathrm{eV}$
Now, the de Broglie wavelength associated with an electron is
$\mathit{\lambda }=\frac{\mathit{h}}{\mathit{p}}$
$\mathit{h}=$ Planck's constant
$\mathit{p}=$ Momentum of electron
$\mathit{m}=$ Mass of electron
$\mathit{p}=\sqrt{2\mathit{m}\mathit{E}}$
$\therefore \mathit{\lambda }=\frac{\mathit{h}}{\mathit{p}}$


$\therefore \mathit{\lambda }=\frac{6.6\times {10}^{-34}\times {10}^{25}}{9.95}=0.66\times {10}^{-9}$
$=6.6{\mathrm{A}}^0$