Question

If the total energy of an electron in a hydrogen atom in excited state is $-3.4eV$, then the de Broglie wavelength of the electron is

• A. $6.6\times 10^{-12}m$
• B. $3\times 10^{-10}m$
• C. $6.6\times 10^{-10}m$
• D. $9.3\times 10^{-12}m$

Answer 1

Answer: (C)

$E_n=\frac{E_1}{n^2}\Rightarrow -3.4eV=-\frac{13.6}{n^2}$

$\Rightarrow n^2=\frac{13.6}{3.4}=4$

$\therefore n=2$

Velocity of electron in $n=2$ is

$v=\frac{2.19\times 10^6}{2}m{s}^{-1}$

$\therefore \lambda =\frac{h}{mv}=\frac{6.6\times 10^{-34}\times 2}{9.1\times 10^{-31}\times 2.19\times 10^6}=6.6\times 10^{-10}m$