Bohr Quantization with linear potential

In summary, using Bohr's quantization rule, we can find the energy levels for a particle in the potential $$U(x) = \alpha\left|x\right|,\alpha>0.$$ By setting the integral of momentum equal to $$2\pi\hbar(n+\frac{1}{2}),$$ we can solve for the energy levels and get an answer of $$E_n = \alpha\left(\frac{3}{2}\frac{\pi\hbar}{\sqrt{2m}}(n+\frac{1}{2})\right)^{2/3}.$$ It is important to note that the integral is only valid up to the turning points at $$\pm \sqrt{\frac{E
  • #1
andre220
75
1

Homework Statement


Using Bohr's quantization rule find the energy levels for a particle in the potential: $$U(x) = \alpha\left|x\right|, \alpha > 0.$$

Homework Equations


##\oint p\, dx = 2\pi\hbar (n + \frac{1}{2})##

The Attempt at a Solution


Okay so:
##\begin{eqnarray}
\oint p\, dx &= \int \sqrt{2m(E-U(x))}\,dx\\
&= \int\limits_{-\infty}^{+\infty} \sqrt{2m(E-\alpha\left|x\right|)}\,dx\\
&= 2\pi\hbar (n+\frac{1}{2}
\end{eqnarray}##
So far, I believe this is correct, but the integral doesn't converge so either I am missing something or I've done something wrong. I can't seem to see what it is. Any help is greatly appreciated.
 
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  • #2
There's a turning point where the square-root becomes imaginary, that's your upper limit...
 
  • #3
Right I see, so I get the two turning points at $$\pm \sqrt{\frac{E^2}{\alpha^2}}$$ and now its just a matter of evalutating the integral.
 
  • #4
Yes, and the integral is easy to carry out.
Also, note that it is symmetric about 0 so you can dispose of the absolute value and take twice the integral from 0 to the turning point.
 
  • #5
Thank you for your help. Got an answer of $$E_n = \alpha\left(\frac{3}{2}\frac{\pi\hbar}{\sqrt{2m}}(n+\frac{1}{2})\right)^{2/3}.$$ which seems okay.
 
  • #6
No problem... But i think you want to put this α inside the parenthesis ;)
 

1. What is Bohr Quantization with linear potential?

Bohr Quantization with linear potential is a theoretical model in quantum mechanics that describes the behavior of a particle in a one-dimensional system with a linear potential. It is an extension of the Bohr model, which only applies to circular orbits in a Coulomb potential.

2. How does Bohr Quantization with linear potential differ from the Bohr model?

The main difference is that the Bohr Quantization with linear potential allows for the particle to move in a straight line, while the Bohr model only allows for circular motion. This makes it more applicable to a wider range of systems, such as a particle in a linear potential well.

3. What is the significance of Bohr Quantization with linear potential in quantum mechanics?

This model helps to explain the quantization of energy levels in a one-dimensional system with a linear potential. It also provides a more accurate description of the behavior of particles in certain systems, such as atoms with a linear potential well.

4. How is Bohr Quantization with linear potential used in practical applications?

This model is used in various fields of physics, such as atomic and molecular physics, to understand and predict the behavior of particles in one-dimensional systems. It is also used in the development of technologies such as quantum computers and nanotechnology.

5. Are there any limitations to Bohr Quantization with linear potential?

Like any other theoretical model, Bohr Quantization with linear potential has its limitations. It does not take into account the effects of relativity and is only applicable to systems with a linear potential. It also does not fully describe the behavior of particles in complex systems, such as molecules with multiple atoms.

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