# If the total energy of an electron in a hydrogen like atom in an excited state is -3.4eV then the debroglie wavelength of the electron is

$\begin{array}{1 1}(a)\;6.6\times 10^{-10}&(b)\;3\times 10^{-10}\\(c)\;5\times 10^{-9}&(d)\;9.3\times 10^{-12}\end{array}$

Total energy=$\large\frac{-e^2}{2r_n}=$$-3.4eV=\large\frac{E_1}{n^2} \therefore n^2=\large\frac{-13.6}{-3.4}$$=4$
$n=2$
Velocity=$\large\frac{u_1}{2}=\frac{2.18\times 10^8}{2}$$cm sec^{-1}$
$\therefore \lambda=\large\frac{h}{mu}$
$\Rightarrow \large\frac{6.6\times 10^{-27}\times 2}{9.108\times 10^{-28}\times 2.18\times 10^8}$
$\Rightarrow 6.6\times 10^{-10}$
Hence (a) is the correct answer.