Particle in a Box: Calculating P from 0 to 0.2 nm

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Homework Help Overview

The problem involves calculating the probability, P, of locating a particle in its lowest energy state within a one-dimensional box, specifically between x = 0 and x = 0.2 nm. The context is rooted in quantum mechanics, particularly the particle in a box model.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the wave function squared to find the probability and question the dimensional consistency of certain factors in the equations. There is also a focus on the use of radians versus degrees in trigonometric functions within the calculations.

Discussion Status

The discussion includes attempts to clarify the mathematical steps involved in the probability calculation. Some participants have provided guidance on the correct use of radians for the sine function, while others express confusion about specific aspects of the calculations.

Contextual Notes

There is an indication that the problem may be considered advanced by some, although it is also noted that it is often included in introductory physics courses. Participants are navigating through potential misunderstandings regarding the mathematical setup and assumptions involved in the problem.

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Homework Statement



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?

Homework Equations



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

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.
 
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ahhppull said:
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

Check the factor highlighted above. Note that this factor should have the dimensions of length.
 
Isn't this advanced physics?
 
lep11 said:
Isn't this advanced physics?

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.).
 
TSny said:
Check the factor highlighted above. Note that this factor should have the dimensions of length.

I still don't understand.

I may have wrote something wrong.
The part that you highlighted should be: (L/n∏)
 
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?
 
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
 
Yes, well done, you're right, radians always for this sort of thing. Took me ages to get used to that.
 
tia89 said:
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

Ok...Thanks!
 

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