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

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The discussion focuses on calculating the probability, P, of locating a particle in a box between x = 0 and x = 0.2 nm in its lowest energy state. The probability is derived from the integral of the wave function squared, ψ², which involves using the sine function. A key point raised is the importance of using radians instead of degrees when calculating the sine function, as the argument must be dimensionless. After correcting the calculation method, the correct probability is confirmed to be 0.05. The conversation emphasizes the need for precision in units and calculations in quantum mechanics.
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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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