Bohr Radius of an atom This is the question I am working on:  The Bohr radius, $a_0$, of the hydrogen atom is the value of $r$ that minimizes the energy $E(r)=\frac{h^2}{2mr^2} - \frac{e^2}{4\pi\epsilon_0r}$ where $h, m,e$, and $\epsilon_0$ are physical constants. Show that $a_0=\frac{4\pi\epsilon_0h^2}{me^2}$ I have tried to plug in $a_0$ for the $r$ value and take the derivative of $E(r)$ but that didn't seem to be getting me any closer to the answer I needed. Any help with this question would be greatly appreciated. Thank you!  
 A: You don't plug in $a_o$ until after having taken the derivative. To find the minimum value of a function you take the derivative and set it equal to zero. So you need to solve the equation $E'(r) = 0$ for $r$. Alternatively, you can take the derivative and then show that $E'(\frac{4\pi\epsilon_0h^2}{me^2}) = 0$.
A: $$\frac{h^2}{2mr^2}-\frac{e^2}{4\pi \epsilon_o r}$$
$$=\frac{h^2}{2m}r^{-2}-\frac{e^2}{4\pi \epsilon_o}r^{-1}$$
By the power rule for derivatives,
$$\frac{d}{dr}(\frac{h^2}{2m}r^{-2}-\frac{e^2}{4\pi \epsilon_o}r^{-1})=-2\frac{h^2}
{2m}r^{-3}+\frac{e^2}{4\pi \epsilon_o}r^{-2}$$
Now to minimize, we have to solve for $E^{'}(r)=0$. Thus,
$$-\frac{h^2}{m}r^{-3}+\frac{e^2}{4\pi \epsilon_o}r^{-2}=0$$
$$\frac{e^2}{4\pi \epsilon_o}r^{-2}=\frac{h^2}{m}r^{-3}$$
$$\frac{e^2}{4\pi \epsilon_o}r=\frac{h^2}{m}$$
$$r=\frac{4\pi \epsilon_0 h^2}{me^2}$$
Therefore,
$$a_0=\frac{4\pi \epsilon_0 h^2}{me^2}$$
A: Note
$$E(r)=\frac{h^2}{2mr^2} - \frac{e^2}{4\pi\epsilon_0r}=\frac{h^2}{2m}\left(\frac1r-\frac{me^2}{4\pi\epsilon_0h^2}\right)^2
-\frac{me^4}{32\pi^2\epsilon_0^2h^2}
$$
which permits minimum energy $E(r)$ at $\frac1r-\frac{me^2}{4\pi\epsilon_0h^2}=0$ and leads to the radius
$$a_0=\frac{4\pi\epsilon_0h^2}{me^2}$$
A: Two things:
The energy that you are defining as $E(r)$ is the total energy and is constant as a function of your coordinate $r$. But, in the way that you wroted, has a little error. 
Let's talk about some theory.
In a circular orbit you can calculate the velocity of the electron from the quantization rule of the angular moment. In fact:
$$ L = m_e v r = \frac{n h}{2  \pi}$$
$n$ is the energy level and $h$ is the Planck constant, wich implies that, in the ground state ($n=1$)
$$ v=\frac{\hbar }{m_e r}$$
$\hbar=h/2\pi$ is the reduced Planck constant. 
The total energy is always defined as
$$ E(r)=\frac{mv^2}{2}  + U(r)$$
But the potential energy in the central force generated by the nucleus is $ - \frac{e^2}{4 \pi \epsilon_0 r} $.
In this way, replacing $v$ and $U$ we have
$$ E(r)=\frac{\hbar^2}{2 m_e r^2} -  \frac{e^2}{4 \pi \epsilon_0 r} $$
I emphasize that this function is constant (consequence of that the electron is in a conservative force field). 
Finally: 
$E(r)=c=E(r)=\frac{\hbar^2}{2 m_e r^2} -  \frac{e^2}{4 \pi \epsilon_0 r}$
Where $c$ is a constant real value.
You can calculate the numerical value of r by taking the derivative of $E(r)$ and then:
$$E'(r)=-\frac{\hbar^2}{m_e r^3} + \frac{e^2}{4 \pi \epsilon_0 r^2} =0$$
Solve for $r$ and the answer is:
$$r=\frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2}$$
And this numerical value is known as the Bohr's radii $a_0$
P.D: Maybe you should post this type of questions in Physics StackExchange.
