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Mattszo
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Homework Statement
Determine an expression for the Bohr radius (a[itex]_{0}[/itex] from the following approximation. The electron moves to the nucleus to lower its potential energy,
V(r) = -[itex]\frac{e^{2}}{r}[/itex]
If the electron is in domain 0[itex]\leq[/itex]r[itex]\leq[/itex][itex]\bar{r}[/itex], then we may write Δp≈[itex]\frac{\hbar}{\bar{r}}[/itex], with corresponding kinetic energy [itex]\frac{\hbar^{2}}{2m\bar{r}^{2}}[/itex]. With this information estimate a[itex]_{0}[/itex] (Bohr radius) by minimizing the total energy.
Homework Equations
Uncertainty ΔpΔx≈[itex]\frac{\hbar}{2}[/itex]
[itex]\bar{r}[/itex] is the Mean of r -------> (I think)
[itex]\hat{H}[/itex][itex]\varphi_{n}[/itex]=E[itex]_{n}[/itex][itex]\varphi_{n}[/itex] or with x-hat and x
For an electron moving with momentum p 2[itex]\pi[/itex]rp=nh
E = K.E + V(x)
[[itex]\hat{A}[/itex], [itex]\hat{B}[/itex]]=[itex]\hat{A}[/itex][itex]\hat{B}[/itex]-[itex]\hat{B}[/itex][itex]\hat{A}[/itex]
[[itex]\hat{A}[/itex], [itex]\hat{B}[/itex]]=0 if they they commute
[[itex]\hat{x}[/itex], [itex]\hat{p}[/itex]]=i[itex]\hbar[/itex]
And I have no idea where this equation came from but I found it in the book and used it for my solution:
[itex]\frac{e^{2}}{r}[/itex]=[itex]\frac{p^{2}_{\vartheta}}{mr^{2}}[/itex]=[itex]\frac{n^{2}\hbar^{2}}{mr^{2}}[/itex] where e is the electric charge of the electron and p[itex]_{\vartheta}[/itex]=mr[itex]^{2}[/itex][itex]\dot{\vartheta}[/itex]
The Attempt at a Solution
E = [itex]\frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}}[/itex]-[itex]\frac{e^{2}}{\bar{r}}[/itex] ... iFrom the "Centripetal Condition" (The equation I got from the book) (r)[itex]\frac{e^{2}}{r^{2}}[/itex]=(r)[itex]\frac{p^{2}_{\vartheta}}{mr^{3}}[/itex]=[itex]\frac{p^{2}_{\vartheta}}{m\bar{r^{2}}}[/itex] ... ii
sub ii into i
E = -[itex]\frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}}[/itex] ... iii
Now from Centripetal Condition:
r=[itex]\frac{n^{2}\hbar^{2}}{me^{2}}[/itex] where p[itex]_{\vartheta}[/itex]=n[itex]\hbar[/itex] ... iv
Put iv into iii
E = -[itex]\frac{p^{2}_{\vartheta}}{2m}[/itex]([itex]\frac{me^{2}}{n^{2}\hbar^{2}}[/itex])[itex]^{2}[/itex]=-[itex]\frac{me^{4}}{2n^{2}\hbar^{2}}[/itex] and let ℝ=[itex]\frac{m_{e}e^{4}}{2\hbar^{2}}[/itex]=13.6eV
Then E[itex]_{n}[/itex]=[itex]\frac{ℝ}{n^{2}}[/itex]
So for n=1, which is the lowest possible state of potential energy, E[itex]_{1}[/itex]=13.6eV
and for n=1 from equation iv
r[itex]_{1}[/itex]=a[itex]_{0}[/itex]=[itex]\frac{n^{2}\hbar^{2}}{me^{2}}[/itex]=5.24x10[itex]^{-9}[/itex]cm=Bohr's radius
The problem is this was too straightforward and simple. I did not correctly use the mean value of [itex]\bar{r}[/itex], 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.
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