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sritter27
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Homework Statement
Instead of electrons in the hydrogen atom experiencing the Coulomb potential, let's say they experience a potential in the form of [tex]V(r) = V_0(\frac{r}{R})^k[/tex], where [tex]V_0 > 0,\: R > 0, \:k > 0[/tex].
With this information find an equation for the radii of the Bohr orbit with quantum number n.
Homework Equations
[tex]V(r) = V_0(\frac{r}{R})^k[/tex]
[tex] L = mvr = n\hbar[/tex]
[tex]F_{Coulomb} = F_{centripetal}[/tex]
The Attempt at a Solution
Under normal circumstances the equation for radii would be easy enough to derive from the fact that [tex]F_{Coulomb} = F_{centripetal}[/tex] for the electrons. So, I attempted to find the derivative of the electron's potential since it should be equal to the negative of Coulomb force. So,
[tex]F_{Coulomb} = -\frac{dV(r)}{dr} = -\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R}[/tex].
Thus,
[tex]-\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{mv^2}{r}[/tex].
And since angular momentum [tex]L = mvr = n\hbar[/tex],
[tex]\frac{n^2\hbar^2}{mr^2} = -\frac{2 k V_0 (\frac{r}{R})^{(k-1)}}{R}[/tex]
But at this point I don't see how I can get r on one side of the equation, so perhaps I approached the problem incorrectly or made a mistake somewhere. Any help will be greatly appreciated.
EDIT: Okay, so my algebra seemed to be off and going back through I think I've come up with a viable answer.
Given as I said before,
[tex]\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{mv^2}{r}[/tex].
[tex]\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{n^2 \hbar^2}{m r^3}[/tex]
[tex]\frac{r^3 k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{n^2 \hbar^2}{m}[/tex]
[tex]\frac{r^{(2 + k)} k V_0}{R^k} = \frac{n^2 \hbar^2}{m}[/tex]
[tex]r^{2 + k} = \frac{n^2 \hbar^2 R^k}{m k V_0}[/tex]
[tex]r_n = \left(\frac{n^2 \hbar^2 R^k}{m k V_0}\right)^{(\frac{1}{2 + k})}[/tex]
So, now I think this is the right answer. If anyone thinks differently, please correct me.
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