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Particle in a Box

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the probability, P, of locating a particle between x = 0 (the left-hand end of
    a box) and x = 0.2 nm in its lowest energy state in a box of length 1.0 nm?

    2. Relevant equations

    Probability = ∫ψ2dx
    ψ = (2/L)1/2sin(n∏x)

    3. The attempt at a solution

    ψ2 = (2/L)sin2(n∏x)
    ∫(2/L)sin2(n∏x)dx = 2/L [x/2 - (L/n∏x)sin(2n∏x/L)] from x = 0 to x = 0.2

    I plugged in the numbers n=1 and L = 1 and got about 0.2.
    The answer is 0.05.
     
  2. jcsd
  3. Mar 2, 2013 #2

    TSny

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    Check the factor highlighted above. Note that this factor should have the dimensions of length.
     
  4. Mar 2, 2013 #3
    Isn't this advanced physics?
     
  5. Mar 2, 2013 #4

    TSny

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    Not necessarily. The particle in a box is often covered in the introductory calculus-based physics course (usually in the 3rd semester of the course in the U.S.).
     
  6. Mar 2, 2013 #5
    I still don't understand.

    I may have wrote something wrong.
    The part that you highlighted should be: (L/n∏)
     
  7. Mar 2, 2013 #6
    I got the answer now, but by changing my calculator to radians when calculating sin(2n∏x/L). Am I suppose to use radian instead of degrees?
     
  8. Mar 2, 2013 #7
    Yes you are... you are working with numbers here, not with angles... in this case the sine is just a function, and want an adimensional argument. While radians are a conventional unit for angles but are not a real units (you call radians to understand that you are speaking of angles but it is still a pure number), degrees are indeed an unit, so you can't use them here
     
  9. Mar 2, 2013 #8
    Yes, well done, you're right, radians always for this sort of thing. Took me ages to get used to that.
     
  10. Mar 2, 2013 #9
    Ok...Thanks!
     
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