1D Particle in a box - locations at set probability

[Nykrus]

Homework Statement

An electron is confined to the region of the x-axis between x = 0 and x = L (where L = 1nm). Given a state n = 3, find the location of the points in the box at which the probability of finding the electron is half it's maximum value

Homework Equations

\psi^2(x)=\frac{2}{L}\sin^2\left(\frac{n\pi x}{L}\right)

The Attempt at a Solution

I understand that the wavefunction squared (above) gives the probability at location x, and its integration gives the probability over set regions between x = 0 and x = l. However, the only way I can see of finding x from a given probability is to assume:

\psi^2(x)=0.5

and try to manipulate the equation to give it in terms of x. Is this the right method? If so, how do I take out the sine term?

Cheers

 

[Ygggdrasil]

The maximum value of ψ²(x) is not 1.

 

[Nykrus]

Ah, I see - half the maximum value. I'm guessing it'd only be 1 if we were considering the entire box, not one point.

Alright, so you determine the maxima by differentiation?

\frac{d\psi^2(x)}{dx}=2\sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{n\pi x}{L}\right)=0

 

[Ygggdrasil]

The rigorous way to determine the maxima would be by differentiation, as you said. However, ψ² has a fairly simple formula and from what you know about sine functions, you should be able to see the maximum value by inspection.

 

[Nykrus]

Hmmm... okay, let's simplify this: 2/L is just a constant, and so's \frac{n\pi}{L}, so that gives us:

\psi^2(x)=A\sin^2\left(ax)

Sine functions go to 1 when ax = 90, and since sin²(ax) is :

\sin^2\left(ax) = \sin\left(ax)\sin\left(ax)

This means... I'm seriously clutching at straws

Hey, I'm a chemist - I'm amazed I've managed this much

 

[Ygggdrasil]

What is the largest value sin(x) can give?

Being a chemist is no excuse for not knowing math (unless of course, you're a biochemist :p)!

 

[Nykrus]

The maximum of sin(x) is 1