Question

A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 eV.
(a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition.(b) Find the wavelength of the emitted radiation.

Answer 1

a) From the energy data we see that the H atom transits from binding energy of $0.85eV$ to excitation energy of $10.2eV$ = Binding energy of -3.4 eV.
So n = 4 to n = 2.

b) We know = $1/\lambda =1.097\times {10}^7(1/4-1/16)$

$⟹\lambda =\frac{16}{1.097\times 3\times {10}^7}$

$=4.8617\times {10}^{-7}=487nm$