Question

If the total energy of an electron in a hydrogen atom in excited state is $-3.4 \,eV$, 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)

$E_{n}=\frac{E_{1}}{n^{2}} \Rightarrow -3.4 eV =-\frac{13.6}{n^{2}}$

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

$\therefore n=2$

Velocity of electron in $n=2$ is

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

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