Discussion Forum

 Consider a hydrogen atom with its electron in the n^th orbital. An electromagnetic radiation of wavelength 90mm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is  (hc= 1242 eV nm)

Answer:

Explanation:

 The kinetic energy of ejected  electron

                   =    Energy of incident photon - energy required to ionize  the electron from nth orbit (all in eV)

      $\therefore$     $10.4=\frac{1242}{90}-|E_{n}|=\frac{1242}{90}-\frac{13.6}{n^{2}}$

                                                (as   $E_{n}\propto \frac{1}{{n^2}}$  and E₁   =-13.6 eV)

             Solving this equation , we get

                                 n=2