Find Bohr Radius from Approximate Potential Energy

[Mattszo]

Homework Statement

Determine an expression for the Bohr radius (a_{0} from the following approximation. The electron moves to the nucleus to lower its potential energy,

V(r) = -\frac{e^{2}}{r}

If the electron is in domain 0\leqr\leq\bar{r}, then we may write Δp≈\frac{\hbar}{\bar{r}}, with corresponding kinetic energy \frac{\hbar^{2}}{2m\bar{r}^{2}}. With this information estimate a_{0} (Bohr radius) by minimizing the total energy.

Homework Equations

Uncertainty ΔpΔx≈\frac{\hbar}{2}
\bar{r} is the Mean of r -------> (I think)
\hat{H}\varphi_{n}=E_{n}\varphi_{n} or with x-hat and x
For an electron moving with momentum p 2\pirp=nh
E = K.E + V(x)
[\hat{A}, \hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}
[\hat{A}, \hat{B}]=0 if they they commute
[\hat{x}, \hat{p}]=i\hbar
And I have no idea where this equation came from but I found it in the book and used it for my solution:
\frac{e^{2}}{r}=\frac{p^{2}_{\vartheta}}{mr^{2}}=\frac{n^{2}\hbar^{2}}{mr^{2}} where e is the electric charge of the electron and p_{\vartheta}=mr^{2}\dot{\vartheta}

The Attempt at a Solution

E = \frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}}-\frac{e^{2}}{\bar{r}} ... i
From the "Centripetal Condition" (The equation I got from the book) (r)\frac{e^{2}}{r^{2}}=(r)\frac{p^{2}_{\vartheta}}{mr^{3}}=\frac{p^{2}_{\vartheta}}{m\bar{r^{2}}} ... ii

sub ii into i

E = -\frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}} ... iii

Now from Centripetal Condition:
r=\frac{n^{2}\hbar^{2}}{me^{2}} where p_{\vartheta}=n\hbar ... iv

Put iv into iii

E = -\frac{p^{2}_{\vartheta}}{2m}(\frac{me^{2}}{n^{2}\hbar^{2}})^{2}=-\frac{me^{4}}{2n^{2}\hbar^{2}} and let ℝ=\frac{m_{e}e^{4}}{2\hbar^{2}}=13.6eV
Then E_{n}=\frac{ℝ}{n^{2}}

So for n=1, which is the lowest possible state of potential energy, E_{1}=13.6eV

and for n=1 from equation iv

r_{1}=a_{0}=\frac{n^{2}\hbar^{2}}{me^{2}}=5.24x10^{-9}cm=Bohr's radius

The problem is this was too straightforward and simple. I did not correctly use the mean value of \bar{r}, because it is a function and I am to find an expression for the Bohr radius.
I used very little of the information provided.
Also I just assumed the lowest energy state, whereas the question says that the electron just goes to a lower energy state.
Also this question is in the chapter where they define the commutator relationships and compatible operators, and really the whole chapter is about that. I find it fishy the solution I provided doesn't use any of that chapters material.I need a new approach to this problem and am out of ideas. Please help.

 

[vela]

The idea here is that the electron is confined to a region of size $\bar{r}$, so applying the uncertainty principle gives you an estimate of the magnitude of its momentum, namely $p \approx \hbar/\bar{r}$, and therefore of its kinetic energy. The total energy of the electron is thus given as a function of $\bar{r}$ by
$$E(\bar{r}) = \frac{\hbar^2}{2m\bar{r}^2} - \frac{e^2}{\bar{r}}.$$ The problem is simply asking you to find the value of $\bar{r}$ for which the total energy of the electron is minimized.

 

[Mattszo]

I see it now, thank you!