Question

If the ionization potential in the ground state for hydrogen is $13.6e.V$. then find the excitation potential of the third orbit.

Answer 1

Ionization potential is $13.6eV$ it means the energy in the ground state will be $E_1=-13.6eV$

Now as we know that the energy in $n^{th}$ prbit is $E_n=\frac{E_1}{n^2}$

So putting $n=3$ for third orbit the energy is $E_3=\frac{-13.6}{9}=1.51eV$

And putting $n=4$ for $fourth$ orbit the energy is $E_4=\frac{-13.6}{16}=-0.85eV$

Thus the excitation potential for third orbit will be $E_4-E_3=(-0.85+1.51)eV=0.66eV$