For the infinite square-well potential, schrodinger eqation

  • Thread starter dawozel
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  • #1
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Homework Statement



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)

Homework Equations


∫ψ^2= Probability


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?
 
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Answers and Replies

  • #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?
 
  • #3
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n should be 4, maybe I'm missing a n in the sine function
 
  • #4
TSny
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n should be 4
What is "n" for the ground state? What is "n" for the first excited state?

maybe I'm missing a n in the sine function
Yes, check that.
 
  • #5
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still not working, might be missing something else
 
  • #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 ?
 
  • #7
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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
 
  • #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?
 
  • #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 = ?
 
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  • #10
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Omg how could i be so blind, thank you! the 4th exited state is actually n=5! thanks the answers coming out correct now
 
  • #11
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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
 
  • #12
TSny
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Omg how could i be so blind, thank you! the 4th exited state is actually n=5! thanks the answers coming out correct now
OK. Good work.
 

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