Question: Is the potential energy of an electron in a hydrogen atom well defined? Is the kinetic energy well defined? Justify your answers. (You need not actually calculate uncertainties.)

The Heisenberg Uncertainty Principle states that the momentum and the position of a particle cannot be measured simultaneously with arbitrary accuracy. For an uncertainty in position $(∆x)$and the uncertainty in momentum $(∆p)$, the uncertainty principle is given as-
$∆x=∆p\ge \frac{h}{2}$

We know that the potential energy of an electron in a hydrogen atom is a function of its position r

$U\left(r\right)=-\frac{e^2}{\left(4\pi {\epsilon }_0\right)r}......\left(1\right)$

However, the position of the electron cannot be defined accurately. According to the Heisenberg uncertainty principle, there is an uncertainty in position,$∆r$ .This uncertainty in position leads to an uncertainty in the potential energy.
$∆U=\frac{e^2}{\left(4\pi {\epsilon }_0\right)r^2}∆r$

Similarly, the kinetic energy is a function of both rotational and the radial momentum operator ${\hat{P}}_{rad}=-i\frac{h}{r}\frac{d}{dr}r$

knowing that the angular momentum takes the values
$L=\sqrt{\mathcal{l}(\mathcal{l}+1)}h$for$\mathcal{l}=0,1,2.....,n-1,$ we have
$\widehat{KE}=\frac{\widehat{{p^2}_{rad}}}{2m}+\frac{L^2}{2m{r}^2}$
$$\begin{gathered} =\frac{1}{2m}{\left(-i\frac{h}{r}\frac{d}{dr}r\right)}^2+\frac{\mathcal{l}(\mathcal{l}+1)h^2}{2m{r}^2} \\ =\underbrace{-\frac{h^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)}+\underset{⏝}{\frac{\mathcal{l}(\mathcal{l}+1)h^2}{2m{r}^2}} \\ K.EradK.Erot \end{gathered}$$

Which agrees with equation (30).

So, we can see the precise definition of the kinetic energy requires the precise value of both the radial momentum and position at the same time, which is not possible, by the virtue of the uncertainty principle.

Therefore, we conclude that neither the potential energy nor the kinetic energy can be well defined for an electron in a hydrogen atom