Question

If the ionisation potential in the ground state for hydrogen is 13.6ev then find the excitation potential of third orbit

Answer 1

Explanation:

Energy required to remove an electron from a gaseous atom or ion is known as ionization energy.

To calculate ionization energy, we use the formula as follows.

          Ionization energy = $E_{\infty} - (-E_{1})$

                      13.6 eV = 0 - $(-E_{1})$

                      $(E_{1})$ = 13.6 eV

Calculate the excitation potential of third orbit as follows.

           Ionization energy = $E_{\infty} - (-E_{3})$    ......... (1)

where,      $E_{3} = \frac{E_{1} \times n^{2}}{z^{2}}$

Here, n = 3 and z = atomic number of hydrogen = 1.

Putting the values in equation (1) as follows.

             Ionization energy = $E_{\infty} - (-E_{3})$  

                                            =  $E_{\infty} - \frac{-E_{1} \times n^{2}}{z^{2}}$

                                            = $0 - \frac{-13.6 eV \times 3^{2}}{1^{2}}$

                                            = 0 - (-122.4 eV)

                                            = 122.4 eV    

Thus, we can conclude that the excitation potential of third orbit is 122.4 eV.