Bohr Quantization with linear potential

  • Thread starter andre220
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  • #1
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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 im missing something or I've done something wrong. I can't seem to see what it is. Any help is greatly appreciated.
 

Answers and Replies

  • #2
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There's a turning point where the square-root becomes imaginary, that's your upper limit...
 
  • #3
75
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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
205
16
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
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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
205
16
No problem... But i think you want to put this α inside the parenthesis ;)
 

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