Question

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 1

Given, the ionisation potential of hydrogen atom = 13.6eV

Hence, the change in energy to remove the electron from n = 2 is given by  

                                $E(n) = -\frac{13.6}{n^2}   eV/atom$

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.