The energy of the $n^{\;\text{th}\;}$ state of Hydrogen atom$E_n = \frac{-13.6}{n^2}\,\mathrm{eV}$For ground state, $n=1$When atom is in first excited state, $n=2$$\therefore \;\; E = \frac{-13.6}{2^2} = -3.4\,\mathrm{eV}$Now, the de Broglie wavelength associated with an electron is$\lambda = \frac{h}{p}$$h=$ Planck's constant$p=$ Momentum of electron$m=$ Mass of electron$\; p = \sqrt{2 m E}$$\therefore \; \lambda = \frac{h}{p}$$\text{Or}\;\; \lambda = \frac{h}{\sqrt{2 m E}}$ $\therefore \;\; \lambda = \frac{6.6 \times 10^{-34} \times 10^{25}}{9.95} = 0.66 \times 10^{-9}$ $= 6.6\,\mathrm{A}^0$