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What is the approximate quantum number of the electron?

  1. Apr 19, 2008 #1

    jk4

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    [SOLVED] particle in a box

    1. The problem statement, all variables and given/known data
    An electron moves with a speed of v = [tex]10^{-4}[/tex]c inside a one-dimensional box of length 48.5nm. The potential is infinite elsewhere. What is the approximate quantum number of the electron?

    2. Relevant equations
    [tex]E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}[/tex]
    n = 1, 2, 3, . . .

    [tex]E = \gamma mc^{2}[/tex]

    3. The attempt at a solution
    I was trying to solve it by finding the total energy of the electron, then using the first equation I stated using the total energy as "En". Then I would try and solve for n but I get a very different number. The answer is supposed to be 4.
     
  2. jcsd
  3. Apr 19, 2008 #2

    Dick

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    n=4 is right. What did you get for the E corresponding v=10^(-4)c? What do you get if you put n=4 into En? BTW v<<c. You could just use (1/2)mv^2 for the energy.
     
  4. Apr 20, 2008 #3
    yes. I got 3.969 for using [tex]E_n=\frac{1}{2}mv^2[/tex].
    Besides, [tex]E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}[/tex] is a result from Schrodinger Equation which is non-relativistic. I am not sure if this form preserve in Dirac Equation or not.
     
  5. Apr 20, 2008 #4

    jk4

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    EDIT:
    ok I got it, thanks for the help.
     
    Last edited by a moderator: Apr 20, 2008
  6. Apr 20, 2008 #5

    Dick

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    Put units on dimensionful numbers ok. I calculate E = (0.5)(9.1095 E -31*kg)(10E-4 x c)^2 and I get 4.09E-22J. We are using the same numbers, unless you are using a different value of c??
     
  7. Apr 20, 2008 #6

    [tex] \frac{1}{2}mv^2 = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}[/tex]

    then

    [tex] n=\frac{mvL}{\pi \hbar}[/tex]

    put [tex] m=9\times 10^{-31}, v=3\times 10^{4}, L=48.5\times 10^{-9} \mbox{ and } \hbar = 1.055\times 10^{-34}[/tex]

    then you can get [tex]n\approx 4[/tex]

    get it?
     
  8. Apr 20, 2008 #7
    wow - this really helped - thanks!
     
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