Ionization energy of a H-atom is 13.6 eV/atom. It requires a photon of energy 1.5 times the minimum which is required to remove the electron. Calculate the wavelength of the emitted electron.

Ionization energy of a H-atom is 13.6 eV/atom. It requires a photon of

1.	Ionization energy of a H-atom is 13.6 eV/atom. It requires a photon of energy 1.5 times the minimum which is required to remove the electron. Calculate the wavelength of the emitted electron.
Answer» the ionisation potential of hydrogen atom = 13.6eV Hence, the change in energy to remove the ELECTRON from N = 2 is given by Since, electron is completely REMOVED from n = 2, so we have the change in energy to be ΔE = E(∞)-E(2) Therefore substituting all the values in the formulae we get, ΔE = 0-\frac{-13.6}{2^2} ΔE =\frac{13.6}{4} eV ΔE = 3.4eV The wavelength of the emitted electron is 3.4eV.