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.
As energy,
E=−13.6n2eV
E1=−13.6eV
E4=13.642=−0.85eV
So, energy of photon is
ΔE=E4−E1=−0.85−(−13.6)=12.75eV
Wavelength corresponding to this energy
λ=hcΔE=6.6×10−34×3×10812.75×1.6×10−19
=0.974×10−7m =974˚A