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Electron in a One Dimensional Infinite Potential Well

  • Thread starter FaraDazed
  • Start date
  • #1
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Homework Statement


An electron is confined to a narrow evacuated tube. The tube, which has length of 2m functions as a one dimensional infinite potential well.

A: What is the energy difference between the electrons ground state and the first excitied state.

B: What quantum number n would the energy difference between adjacent energy levels be 1eV - which is measurable, unlike the result of part A.


Homework Equations


[/B]
[tex]
E_n=\frac{\hbar^2 \pi^2 n^2}{2mL^2}
[/tex]


The Attempt at a Solution


I must not be using the equation correctly, as I am getting values many many orders of magnitudes out. Is m in the equation the mass of the particle? So in this case the mass of the electron?


[tex]
E_1=\frac{\hbar^2 \pi^2 n^2}{2mL^2}= \frac{(1.055 \times 10^{-34})^2 \pi^2 }{2(9.109 \times 10^{-31})2^2} = 1.507 \times 10^{-38} J \\
E_2=\frac{\hbar^2 \pi^2 n^2}{2mL^2}= \frac{(1.055 \times 10^{-34})^2 \pi^2 }{2(9.109 \times 10^{-31})} = 6.03 \times 10^{-38} J \\
\Delta E = E_2 - E_1 = 4.523 \times 10^{-38} J = 2.82 \times 10^{-19} eV
[/tex]

For B: I got a quantum number of around 3.5 billion :D, so I have to be doing something wrong. Also my main question for part B was the bit where it carries on with "- which is measurable, unlike the result of A", I have no idea what they mean by that, so if anyone does I would really appreciate it.

Thanks :)
 
Last edited:

Answers and Replies

  • #2
34,370
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The answer looks good. 2 meters is just an extremely wide potential well.
An energy difference of 1 eV is measurable, an energy difference of 10-19 eV (for an electron) is not.
 
  • #3
347
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The answer looks good. 2 meters is just an extremely wide potential well.
An energy difference of 1 eV is measurable, an energy difference of 10-19 eV (for an electron) is not.
Oh right, ok thanks. Because part B was talking about 1eV I thought my answer was way way off.

And thanks for clarifying the wording on part B, that makes sense. Thanks :)
 
  • #4
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I am really stuck on part B now, I just cant see how it can be solved. (Im using the other version of the equation with h instead of h-bar)

[tex]
(1.602 \times 10^{-19})=\frac{h^2n_1^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2}
[/tex]

So obviously the RHS has to be equal to the charge of the electron, but I just cannot see how I can solve this. Any help/advice/feedback is greatly appreciated.

Thanks :)
 
  • #5
34,370
10,445
Don't forget the units.

You have calculated the energy levels already, for adjacent energy levels n1 is just 1 larger than n2 (using your sign convention). This is a quadratic equation.
 
  • #6
347
2
Don't forget the units.

You have calculated the energy levels already, for adjacent energy levels n1 is just 1 larger than n2 (using your sign convention). This is a quadratic equation.
I can t seem to get a quadratic, as the way I am doing it the n^2 cancel. What I have done is below

[tex]
1eV=\frac{h^2n_1^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
1eV=\frac{h^2(n_2+1)^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
1eV=\frac{h^2(n_2^2+2n_2+1)}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
1eV=\frac{h^2(n_2^2+2n_2+1)8mL^2-h^2n_28mL^2}{8mL^2} \\
(1.602 \times 10^{-19})8mL^2=h^2(n_2^2+2n_2+1)8mL^2-h^2n_28mL^2 \\
(1.602 \times 10^{-19})8mL^2=h8mL^2n_2^2+2h^28mL^2n_2+h^28ml^2-h^2n_28mL^2 \\
(1.602 \times 10^{-19})8mL^2=2h^28mL^2n_2 + h^28mL^2\\

n_2=\frac{(1.602 \times 10^{-19})h^28ml^2-h^28mL^2}{2h^2mL^2}
[/tex]

Which if I have calculated it correctly comes out to negative one half, which is impossible.

EDIT: After using wolfram alpha, and sorting out an error, I got the result of n to be equal to the charge on the electron, which is also wrong.
 
Last edited:
  • #7
34,370
10,445
Ah right, the equation becomes linear. Well, just makes it easier.

Something with the 8mL^2 went wrong in the fourth line.
 

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