Probability of finding a particle given psi squared graph

[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.

 

[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?

 

[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...

 

[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

?

 

[Linus Pauling]

Nevermind got em