Probability of finding a particle given psi squared graph

1. Apr 30, 2010

Linus Pauling

1.
The figure shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a 2.40×10−2-mm-wide strip at:

I'm then asked the probability of finding a particle at various spots on the x-axis. We'll go with x = 0.0 here.

2. P(x)*delta(x) = psi2*delta(x)

3. The big triangle has area 0.5(2mm)(0.5) = 0.5mm2. I then integrated from -1.2*10-2 to +1.2*10-2 because that width is centered around the 0.0 point of interest. I then evaluated the solution, 0.5x, at the boundaries, obtaining an answer of 0.012... although I don't think I'm going about this quite right.

2. Apr 30, 2010

Linus Pauling

I know that P will be psi(x) squared multiplied about the length L given in the problem. However, from the graph I can get the integral of psi squared, which is simply the area under the curve. How do I compute psi squared itself?

3. Apr 30, 2010

Linus Pauling

I know this is probably simple but I just don't see it. I don't see how to get an equation that's a function of x that I can plug my x values into. For example, when it asks me for the probability at x=o over a length L, I know that the integral of psi squared is just (1/2)bh = 0.5, but it can't be that I just multiply that by L because the position x didn't play a role................

4. Apr 30, 2010

Linus Pauling

Really lost here...

I just tried solving it using a problem in my book as an analogy. For the probability at x = 0:

I said that the equation describing one half of the "big triangle" is 0.50(1-x/1nm), which is psi squared. Solving with x = zero then multiplying by two to account for the otehr half of the triangle, I get 0.5*2 = 1, which I multiplied by 2.4*10^-2 = 0.024

???

5. Apr 30, 2010

Linus Pauling

Nevermind got em