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Question

The ionisation energy of the hydrogen atom is given to be 13.6eV. A photon falls on a hydrogen atom which is initially at the ground state and excites it to the (n=4). Calculate the wavelength of the photon.

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Solution

As energy,
E=13.6n2eV

E1=13.6eV

E4=13.642=0.85eV
So, energy of photon is
ΔE=E4E1=0.85(13.6)=12.75eV

Wavelength corresponding to this energy
λ=hcΔE=6.6×1034×3×10812.75×1.6×1019

=0.974×107m =974˚A

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