Discussion Forum



1)

 Consider a hydrogen atom with its electron in the nth 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)


A) 2

B) 3

C) 1

D) 4

Answer:

Option A

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