Using Bohr quantization rules to calculate energy levels

  • #1

Homework Statement


Use the Bohr quantization rules to calculate the energy levels for a particle moving in a potential
\begin{equation}
V\left(r\right)= V_0 \left(\frac{r}{a}\right)^k
\end{equation}
where k is positive and very large. Sketch the form of the potential (done) and show that the energy values approach $E_n = kn^2$, where k is a constant.

Homework Equations


\begin{equation}
L=mvr=n\hbar
\end{equation}

The Attempt at a Solution


So, first I derived an equation for the force
\begin{equation}
-\vec{\nabla} V = -\frac{kV_0}{a}\left(\frac{r}{a}\right)^{k-1} \hat{r} = \vec{F}\left(r\right)
\end{equation}

Setting the force equal to the centripetal force give me:
\begin{align}
-\frac{kV_0}{a}\left(\frac{r}{a}\right)^{k-1} &= m\frac{v^2}{r} \\
&= \frac{n^2\hbar^2}{mr^3}
\end{align}
Where the centripetal force is rewritten in terms of $n\hbar$ via the quantization rule. Now with a inverse square force, one can easily solve for r, but in this case, after algebraic manipulation, I have
\begin{align}
r^{k+2}&= - \frac{n^2\hbar^2 a^{k+2}}{kV_0} \\
r_n &= \sqrt[k+2]{- \frac{a^k n^2\hbar^2}{kV_0}}
\end{align}
Now, here I have a problem I have no constraints to prevent a complex radius, because $n, \hbar,$ and $k$ are all positive, and $V_0$ or $a$ are not specified in any way.

I have kept the exponents exact, but since k is positive and very large, I'd think $k \simeq k+2 \simeq k-1$, but I haven't used this. Is this reasonable? i would think $x^{1000} \simeq x^{1001}$, etc. . .

So the next step would be to use the quantized orbits to derive the energy,
\begin{align}
E &= \frac {p^2}{2m} + V \\
&=-\frac{kV_0}{2}\left(\frac{r}{a}\right)^k + V_0 \left(\frac{r}{a}\right)^k \\
&= \left(1-\frac{k}{2}\right)V_0\left(\frac{r}{a}\right)^k
\end{align}

Now, if I allow the assumption that $k \simeq k+2$, I could substitute $r_n$ for $r$, and get:
\begin{align}
E &= \left(1-\frac{k}{2}\right)V_0\frac{-n^2\hbar^2 a^k}{kV_0 a^k} \\
&= \left(1-\frac{k}{2}\right)\frac{1}{k}\hbar^2 n^2 \\
\end{align}
We're given that k is very large and positive, which reduces the former equation to
\begin{equation}\frac{1}{2}\hbar^2 n^2
\end{equation}

So this meets the target of $E_n = kn^2$, where $k= \frac{\hbar^2}{2}$. Does this look like a reasonable quantization? Should the energy be positive? Any constructive criticism, etc. would be greatly appreciated. Cheers!
 

Answers and Replies

  • #2
DrClaude
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Setting the force equal to the centripetal force give me:
\begin{align}
-\frac{kV_0}{a}\left(\frac{r}{a}\right)^{k-1} &= m\frac{v^2}{r} \\
&= \frac{n^2\hbar^2}{mr^3}
\end{align}
The forces have to be equal and opposite. Another way of saying it is that the sum of forces should be zero.

\begin{align}
r^{k+2}&= - \frac{n^2\hbar^2 a^{k+2}}{kV_0} \\
r_n &= \sqrt[k+2]{- \frac{a^k n^2\hbar^2}{kV_0}}
\end{align}
There is an error in the RHS on the first line, where you should have ##a^k##, but I guess it is a typo since the next line is correct.

Now, here I have a problem I have no constraints to prevent a complex radius, because $n, \hbar,$ and $k$ are all positive, and $V_0$ or $a$ are not specified in any way.
The first part of the question asks you to plot the potential. Didn't you need to assume whether ##V_0## and ##a## are positive or negative? What would change if you were to change their sign?



\begin{align}
E &= \left(1-\frac{k}{2}\right)V_0\frac{-n^2\hbar^2 a^k}{kV_0 a^k} \\
&= \left(1-\frac{k}{2}\right)\frac{1}{k}\hbar^2 n^2 \\
\end{align}
Careful with the signs here (although this might not be relevant once you take into account what I wrote above).

Should the energy be positive?
What do you think, considering the form of the potential?
 
  • #3
My mistake was using a minus sign in my force; the centripetal force should have had a minus sign too. I used $\hat{r}$ for my force, but not with the centripetal force, which would also be negative. I should have just used the magnitude. The exponent k isn't specified as odd or even, so we really have 4 cases (depending on the sign of a), which is making the problem much harder than it needs to be. However, one would assume that we want units to vanish (since $V_0$ is a potential energy and $\left(\frac{r}{a}\right)^k$ would be meters to the kth power, which is meaningless), so I think it would be a radial distance, which is always positive. $V_0$ is not given, so I'm assuming it's positive. With a very high, positive exponent, it's basically parabolic in shape (well it slowly creeps up to $V_0$ at $a$, then grows exponentially). The sign on V just points it up or down. I would assume we want a positive potential, since it would lead to an attractive force, and we're considering orbital energy states (Bohr quantization). I've got it now, I just have to write it out again.

Thanks for trying to help mate! I do appreciate it!
 
  • #4
DrClaude
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7,647
4,094
However, one would assume that we want units to vanish (since $V_0$ is a potential energy and $\left(\frac{r}{a}\right)^k$ would be meters to the kth power, which is meaningless), so I think it would be a radial distance, which is always positive.
Indeed, the role of a is only to scale the radial distance to make it dimensionless, so you can assume it is positive.

$V_0$ is not given, so I'm assuming it's positive. With a very high, positive exponent, it's basically parabolic in shape (well it slowly creeps up to $V_0$ at $a$, then grows exponentially). The sign on V just points it up or down. I would assume we want a positive potential, since it would lead to an attractive force, and we're considering orbital energy states (Bohr quantization).
Exactly. The problem doesn't make sense if ##V_0 \leq 0 ##.
 
  • #5
Still getting a positive energy, if anyone sees any flaws in my methodology please post! Thanks & cheers.
 
  • #6
DrClaude
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Still getting a positive energy, if anyone sees any flaws in my methodology please post!
But the energy should be positive. Kinetic energy is always positive, and the potential energy is everywhere > 0.
 
  • #7
But for a bound system energy should be negative, like the electron in the hydrogen atom. You do the same thing, but with Coulomb Force. The potential for that force is negative, so it is more than the kinetic energy term and you're left with negative energy which is ##-\frac{R}{n^2}##, where ##R## is the Rydberg constant ##\frac {m_e}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2 = 13.6 eV##.
 
  • #8
DrClaude
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You get a negative energy because the potential is negative: ##V(r) = - e^2 / 4 \pi \epsilon_0 r##. And even then, it is purely arbitrary, since the potential is negative only because it is chosen to be 0 at infinity. We could choose the potential such that it is 0 at ##r=0##, or such that the ground state of hydrogen would have an energy of 0, and nothing would change except for a physically irrelevant constant added to the energy.
 
  • #9
I think I see what you mean. I was worried because I remember a professor teaching that a system with negative energy indicates a bound system. Also, because conservative forces must satisfy ##-\vec{\nabla}V=\vec{F}##, I thought the sign surely must not be arbitrary (although one can naturally setup the coordinate system anyway one wishes). I suppose this is just inevitable since it's not an inverse square law. I guess this is what I get for being more interested in a Lagrangian Field Density rather than a Lagrangian. Thanks for bearing with me!
 
  • #10
DrClaude
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I think I see what you mean. I was worried because I remember a professor teaching that a system with negative energy indicates a bound system.
That statement is only correct when we choose ##E=0## when the constituents of the system are infinitely apart (which, admitedly, is often the case in QM).

Also, because conservative forces must satisfy ##-\vec{\nabla}V=\vec{F}##, I thought the sign surely must not be arbitrary (although one can naturally setup the coordinate system anyway one wishes).
The coordinate system has nothing to do with it. The negative sign just ensures that the forces acts in the direction where the potential energy will decrease, where the gradient will be negative. And remember that you get the same force whichever constant ##C## we choose to add to the potential, ##-\vec{\nabla}(V + C) = -\vec{\nabla}V##

I suppose this is just inevitable since it's not an inverse square law.
It is not related to the form of the potential. All known laws of physics are independent of the choice of the zero of energy.
 
  • #11
I don't think I am explaining myself clearly, but I am good now. Thanks!
 

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