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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
Identify the quantum numbers n of the upper and the lower energy states involved in the transition.

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Solution

Initially, the electron is at n1=4 shell
As, 13.6n2=B.E=0.85,n1=4
After deexcitation, it is at n2=2 shell.
As,
Excitation Energy = 13.613.6n22=10.2eV
So, the wavelength of emitted photon is,
heλ=[13.6n2213.6n21]1.6×1019
λ=6.6×1034×3×108×161.6×1019×13.6×3
λ=4852˚A

1112444_1115448_ans_a771495b58d1484d82645ef222d1766b.jpg

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