Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy level of the hydrogen atom is - 13.6 eV.
[2020 | Delhi | Section - C | SET - 1 SET - 2 Q.29 Q.31 ]Ans : Energy of the electron in first excited state i.e. n=2
E2=[-13.6/(2)2] eV
E2=-3.4 eV
KE=-TE
KE=-E2
KE=-(-3.4)eV
KE=3.4eV
KE=(3.4e) eV
K=(3.4e) eV
de Broglie Wavelength
λ=h/√(2mK)
λ=[6.6×10-34]/[√(2×9.1×10-31×3.4×1.6×10-19)]
λ=0.663×10-9m
λ=0.663 nm
λ=6.63 Å
