The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Solution

In Bohr's model, as per the quantisation of angular momentum we have,

mvr = $\frac{nh}{2 π}; and$

$\frac{{mv}^{2}}{r} = \frac{{Ze}^{2}}{4 {πε}_{0} r^{2}}$

which gives
Kinetic energy, $E_{k} = \frac{1}{2} {mv}^{2} = \frac{{Ze}^{2}}{8 {πε}_{0} r}; r = \frac{4 {πε}_{0} h^{2}}{{Ze}^{2} m} n^{2} .$
Here, we have $\frac{1}{4 {πε}_{o}} = K$
$∴$ Kinetic energy, K.E = $\frac{{Kze}^{2}}{2 r}$

These relations have nothing to do with choice of the zero of potential energy.

Now, choosing the zero of potential energy at infinity, we have
$E_{p} = \frac{- {Ze}^{2}}{4 {πε}_{0} r} = - \frac{k Z e^{2}}{r} which gives, E_{p} = - 2 E_{k}$
E_p is the potential energy.

Thus, total energy, $E = E_{k} + E_{P} = - E_{k}$

(a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity.
From the above result of E = – E_k, the kinetic energy of electron in this state is + 3.4 eV.

(b) Using E_p = – 2 E_k, potential energy of the electron is – 2 × 3.4 eV = – 6.8 eV.

(c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV. This is independent of the choice of the zero of potential energy.

The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen.