Using the force law to obtain total energy of an electron as a function of its radius I am working on a problem which starts saying determine the total energy of a hydrogen atom with an electron moving with momentum $p$ at a radius $r$. 
For that part I got:
$E = \frac{p^2}{2m_e} - \frac{e^2}{r}$
Which is just the kinetic energy plus the potential energy. If I didn't goof big time that should be good to go. 
But now it asks:
"Use the force law to obtain the total energy as a function of radius. What radius corresponds to the lowest possible energy?" and I am completely lost.
What force law is it asking for here?
If it's worth noting, this is a problem trying to get you to understand the failure of classical mechanics at quantum levels.
 A: If the orbit is circular, then $p=\rm{const}$ and $r=\rm{const}$. $E$  is constant and negative (for a bound state) even though the orbit is not circular. So, one can determine $r$ from this equation: $$r=\rm{e}/\left(p^2/2m_e-E\right).$$
The minimum is zero (no kinetic energy, only the negative potential one), which is not supported experimentally.
A: The force law in question here is:
$$F = \frac {mv^2}{r}$$
Since we know that this centripetal force comes entirely from the electrostatic force we get this
$$F = \frac {mv^2}{r} = \frac {e^2}{r^2}$$
$K$ is excluded from the right hand side force through the use of convenient units for charge. By manipulating a bit
$$p = mv \xrightarrow {} p^2 = {m^2v^2}$$
$$E_{kinetic} = \frac {p^2}{2m}$$
$$E_{kinetic} = \frac {m^2v^2}{2m} = \frac {mv^2}{2}$$
From above
$$mv^2 = \frac {e^2}{r}$$
so
$$\frac{mv^2}{2} = \frac{e^2}{2r} = E_{kinetic}$$
Plugging back into the original formula for the total energy
$$E_{total} = E_{kinetic} - E_{potential} = \frac{e^2}{2r} - \frac{e^2}{r} = -\frac{e^2}{2r}$$
For all future Googlers this is from Introduction to Quantum Mechanics French and Taylor 2-16 part B
