In an excited state of hydrogen like atom an electron has a total energy of -3.4 eV. Then its de Broglie wavelength $\mathrm{\lambda }$ is nearly

$\begin{array}{l}\mathrm{E}=-13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2};-3.4=-13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}\\ {\left(\frac{\mathrm{z}}{\mathrm{n}}\right)}^2=\frac{3.4}{13.6}=\frac{1}{4};\therefore \frac{\mathrm{z}}{\mathrm{n}}=\frac{1}{2}\Rightarrow \mathrm{z}=\frac{\mathrm{n}}{2}\to 1\end{array}$
$\begin{array}{l}\to {\mathrm{\lambda }}_{\mathrm{n}}=\frac{2{\mathrm{\pi r}}_{\mathrm{n}}}{\mathrm{n}}=\frac{2\mathrm{\pi }}{\mathrm{n}}\left(0.53\times \frac{{\mathrm{n}}^2}{\mathrm{z}}\right){\mathrm{A}}^{\mathrm{o}}\\ =\frac{2\mathrm{\pi }}{\mathrm{n}}\left(0.53\times \frac{{\mathrm{n}}^2}{\mathrm{n}/2}\right){\mathrm{A}}^{\circ };=4(3.14)(0.53){\mathrm{A}}^0\\ =6.6{\mathrm{A}}^0=6.6\times {10}^{-10}\mathrm{m}.\end{array}$