# Particle in a Box

1. Mar 2, 2013

### ahhppull

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. Mar 2, 2013

### TSny

Check the factor highlighted above. Note that this factor should have the dimensions of length.

3. Mar 2, 2013

### lep11

Isn't this advanced physics?

4. Mar 2, 2013

### TSny

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.).

5. Mar 2, 2013

### ahhppull

I still don't understand.

I may have wrote something wrong.
The part that you highlighted should be: (L/n∏)

6. Mar 2, 2013

### ahhppull

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?

7. Mar 2, 2013

### tia89

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

8. Mar 2, 2013

### Rooted

Yes, well done, you're right, radians always for this sort of thing. Took me ages to get used to that.

9. Mar 2, 2013

Ok...Thanks!