Given the ground state energy E0=−13.6eV and Bohr radius a0=0.53oA . 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.
rn=a0n2Zrn=0.53n2Z
when the electron is in its ground state:
r1=0.531Z=0.53(Z=1forhydrogenatomforwishE∘=−13.6e
r2=0.53×(2)2=0.53×4Fromde−Broglie′srelation:2πrn=nλλ=2πrnn1=2π×0.531=2π×0.53λ′=2πr2n2=2π×0.53×42=2π×0.53×2=2λ
∴ Wavelength would get doubled.