How Do Modified Potentials Affect Bohr Orbit Radii?

[sritter27]

Homework Statement

Instead of electrons in the hydrogen atom experiencing the Coulomb potential, let's say they experience a potential in the form of $$V(r) = V_0(\frac{r}{R})^k$$, where $$V_0 > 0,\: R > 0, \:k > 0$$.

With this information find an equation for the radii of the Bohr orbit with quantum number n.

Homework Equations

$$V(r) = V_0(\frac{r}{R})^k$$

$$L = mvr = n\hbar$$

$$F_{Coulomb} = F_{centripetal}$$

The Attempt at a Solution

Under normal circumstances the equation for radii would be easy enough to derive from the fact that $$F_{Coulomb} = F_{centripetal}$$ 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,
$$F_{Coulomb} = -\frac{dV(r)}{dr} = -\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R}$$.
Thus,
$$-\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{mv^2}{r}$$.

And since angular momentum $$L = mvr = n\hbar$$,
$$\frac{n^2\hbar^2}{mr^2} = -\frac{2 k V_0 (\frac{r}{R})^{(k-1)}}{R}$$

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,
$$\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{mv^2}{r}$$.

$$\frac{k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{n^2 \hbar^2}{m r^3}$$

$$\frac{r^3 k V_0 (\frac{r}{R})^{(k-1)}}{R} = \frac{n^2 \hbar^2}{m}$$

$$\frac{r^{(2 + k)} k V_0}{R^k} = \frac{n^2 \hbar^2}{m}$$

$$r^{2 + k} = \frac{n^2 \hbar^2 R^k}{m k V_0}$$

$$r_n = \left(\frac{n^2 \hbar^2 R^k}{m k V_0}\right)^{(\frac{1}{2 + k})}$$

So, now I think this is the right answer. If anyone thinks differently, please correct me.

 

[Jhero]

Thank you for your time.

Your approach seems correct, and your final equation for the radius of the Bohr orbit with quantum number n looks correct as well. Good job!