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For the infinite square-well potential, schrodinger eqation

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    For the infinite square-well potential, find the probability that a particle in its fourth excited state is in each third of the one-dimensional box:
    (0 to L/3)
    (L/3 to 2L/3)
    and (2L/3 to L)

    2. Relevant equations
    ∫ψ^2= Probability


    3. The attempt at a solution
    So from ∫ψ^2 for the first third of the problem i got that my probability equation should be
    1/3-(1/((2πn))(sin(2π/3)) - 0 where n is that excited state but I am getting a wrong answer of about .10 off , is there something wrong with my equation?
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    TSny

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    Hello, dawozel.

    Shouldn't there be an "n" in the argument of the sine function?

    Just to check: what is the value of "n" for the 4th excited state?
     
  4. Dec 8, 2013 #3
    n should be 4, maybe I'm missing a n in the sine function
     
  5. Dec 8, 2013 #4

    TSny

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    What is "n" for the ground state? What is "n" for the first excited state?

    Yes, check that.
     
  6. Dec 8, 2013 #5
    still not working, might be missing something else
     
  7. Dec 8, 2013 #6

    TSny

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    What value are you getting for the probability of the particle to be in the range 0 < x < L/3 ?
     
  8. Dec 8, 2013 #7
    Well I've been testing this out with an example I Know the answer to (where N=3) but I'm getting an answer of .306 when it should be .299
     
  9. Dec 8, 2013 #8

    TSny

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    Hmm. The probability for the particle to be between x = 0 and x = L/3 when n = 3 is exactly 1/3.

    Did you decide whether or not there should be a factor of n in the argument of the sine function in your result of

    ##\frac{1}{3} - \frac{1}{2 \pi n} \sin(\frac{2 \pi}{3})##?

    Is your calculator in radian mode when you make the calculations?

    Can you show the formula that you used with the numbers plugged in for the case of n = 3?
     
  10. Dec 8, 2013 #9

    TSny

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    I think there is still a misunderstanding of what the value of n should be for the fourth excited state.

    Can you specify the value of n for each of the following?

    (1) Ground state. n = ?
    (2) First excited state. n = ?
    (3) Second excited state. n = ?
     
  11. Dec 9, 2013 #10
    Omg how could i be so blind, thank you! the 4th exited state is actually n=5! thanks the answers coming out correct now
     
  12. Dec 9, 2013 #11
    For reference the necessary function to calculate this is

    1/A - 1/(2πN) x Sin(2πN/3)

    Where N is the state
    A is the amount of the box you want
    Just plug in your limits and solve
     
  13. Dec 9, 2013 #12

    TSny

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    OK. Good work.
     
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