Question

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

Answer 1

In Bohrs model, mvr = nh and $\frac{m{v}^2}{r}=\frac{Z{e}^2}{4\pi {\epsilon }_0{r}^2}$, which give,
$T=\frac{1}{2}m{v}^2=\frac{Z{e}^2}{8\pi {\epsilon }_{0}r};r=\frac{4\pi {\epsilon }_0{h}^2}{Z{e}^{2}m}{n}^2$
These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity, we have $V=-\left(Z{e}^2/4\pi {\epsilon }_{0}r\right)$, which gives $V=-2T$ and $E=T+v=-T$.

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

(b) Using V = 2T, potential energy of the electron is = 6.8 eV.

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