The energy of the state of Hydrogen atom
$={\mathrm{E}}_{\mathrm{n}}=\frac{-13.6}{{\mathit{n}}^2}$
For ground state, $\mathrm{n}=1$
When atom is in second excited state, $\mathrm{n}=3$
$\therefore \mathrm{E}=\frac{-13.6}{3^2}-1.51\mathrm{eV}$
Now, the de Broglie wavelength associated with an electron is
$\mathit{\lambda }=\frac{\mathit{h}}{\mathit{p}}$
$\mathit{h}=$ Planck's constant
$\mathrm{p}=$ Momentum of electron
$\mathrm{m}=$ Mass of electron
$\mathit{p}=\sqrt{2\mathit{m}\mathit{E}}$
$\therefore \mathit{\lambda }=\frac{\mathit{h}}{\mathit{p}}$


$\therefore \mathit{\lambda }=\frac{6.6\times {10}^{-34}\times {10}^{25}}{6.631}$
$=0.995\times {10}^9\mathrm{m}$
$=9.95{\mathrm{A}}^{\circ }$