Question

What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?

Answer 1

In ground state, the potential energy of a hydrogen atom is zero.
An electron is bound to the nucleus with an energy of 13.6 eV.
Therefore, we have to give 13.6 eV energy to move the electron from the nucleus.
Let us calculate the excitation energy required to take an atom from the ground state (n = 1) to the first excited state (n = 2).
$E=13.6\times \left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)\mathrm{eV}$
Therefore, the excitation energy is given by

$$\begin{gathered} E=13.6\times \left(\frac{1}{1^2}-\frac{1}{2^2}\right)\mathrm{eV} \\ E=13.6\times \frac{3}{4}\mathrm{eV}=10.2\mathrm{eV} \end{gathered}$$
Energy of 10.2 eV is needed to take an atom from the ground state to the first excited state.

∴ Total energy of an atom in the first excitation state = 13.6 eV + 10.2 eV = 23.8 eV