# Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapter

1. Jan 28, 2012

### Mattszo

1. The problem statement, all variables and given/known data
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$\leq$r$\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.

2. Relevant 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$\pi$rp=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}$

3. 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.

Last edited: Jan 28, 2012
2. Jan 29, 2012

### vela

Staff Emeritus
Re: Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapt

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.

3. Jan 29, 2012

### Mattszo

Re: Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapt

I see it now, thank you!