Solving Infinite Square Well: Find Probability of Electron in 0.15nm

[khemist]

Homework Statement

An electron is trapped in a 1.00 nm wide rigid box. Determine the probability of finding the electron within 0.15nm of the center of the box (on either side of the center) for a) n = 1

Homework Equations

Int[-0.15nm, 0.15nm] psi^2 dx

The Attempt at a Solution

I solved the integral fairly easily, but I got a value of 4.2% for my probability. Later in the question, it asks to compare this value to the expected value classically. Classically, in a 1nm wide box over a distance .3nm one would expect the probability to be 30%. Does this seem to make any sense? I talked to my teacher and he thought the values came out a little weird.

 

[vela]

I got about 56% for the probability. I suspect you're using the wrong combination of wave function and limits. Show us your calculations in detail if you can't figure it out.

 

[khemist]

My limits were from -0.15nm to 0.15nm. L = 1.00 nm.

psi = root(2/L)sin(nx(pi)/L)

psi^2 = 2/L sin^2(nx(pi)/L)

My solution to the integral (using -a = -.015nm and a = 0.15nm) was 2/L(x/2 (from -a to a) - L/(2n(pi))sin(2nx(pi)/L)) from -a to a. (the a/2 and other side of the minus sign were separate integrals, using the (1-cos2x)/2 = sin^2(x))

Also, does anyone have a link to I can learn how to type this up using latex or whatever language to have formal equations written?

 

[vela]

Your wave function is for a potential which is zero between x=0 and x=L. You either need to change the limits on your integral, or use the wave function for the potential that's centered about x=0.

You can start with https://www.physicsforums.com/showthread.php?t=386951 to see how to format equations using LaTeX in this forum.

 

[khemist]

So the wave function I have, sqrt(2/L)sin(n(pi)x/L), is not centered about the origin? Would that mean I must compensate with a limit change?

 

[vela]

Yup, or keep your current limits and use the wave function that is centered at the origin.