Question

Given the ground state energy $E_0=-13.6eV$ and Bohr radius $a_0=0.53\stackrel{o}{A}$ . Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.

Answer 1

$r_n=a_0\frac{n^2}{Z}{r}_n=0.53\frac{n^2}{Z}$

when the electron is in its ground state:

$r_1=0.53\frac{1}{Z}=0.53(Z=1forhydrogenatomforwish{E}_{\circ }=-13.6e$

$r_2=0.53\times {\left(2\right)}^2=0.53\times 4Fromde-Brogli{e}^{\prime }srelation:2\pi r_n=n\lambda \lambda =\frac{2\pi r_n}{n_1}=\frac{2\pi \times 0.53}{1}=2\pi \times 0.53{\lambda }^{{}^{\prime }}=\frac{2\pi r_2}{n_2}=\frac{2\pi \times 0.53\times 4}{2}=2\pi \times 0.53\times 2=2\lambda$

$\therefore$ Wavelength would get doubled.