# Quantum Mechanics - Finite Square Well - Graphical Solution

1. Apr 10, 2013

### amr55533

1. The problem statement, all variables and given/known data

http://img842.imageshack.us/img842/4917/physp6.jpg [Broken]

I am trying to solve the above problem. However, I am supposed to solve it with the following values:

U=54.7eV
L=0.2nm

Particle is an electron, so:

m=9.109E-13kg=0.511eV/c^2

Essentially I am supposed to find the conditions for which no solution is possible using a graphical method. This is where I am stuck. The book that I have does not discuss the graphical method, so I am not sure where to begin with it or where the values come in to play. I have solved it using a traditional method seen below:

2. Relevant equations

Schrodinger Equation

3. The attempt at a solution

http://img834.imageshack.us/img834/2686/physw6.png [Broken]

As can be seen above:

k=√[(2mE)/h^2]
α=√[(2m(U-E))/h^2]

kLcsc(kL)=√[(2mUL^2)/h^2]

since kLcsc(kL)=kL/sin(kL) can never be smaller than one, (2mUL^2)/h^2 < 1

Also, since sin(kL) is in the denominator, no solutions would be possible when kL is equal to zero or a multiple of π.

To do it graphically, would I simply draw a graph of kLcsc(kL) and show where it is undefined? I am confused. Also where would the values I was given come into play?

Thanks!

Last edited by a moderator: May 6, 2017
2. Apr 10, 2013

### vela

Staff Emeritus
I didn't check your work completely, but it looked okay. You can rewrite that last expression as
$$\sin kL = c(kL)$$ where $c=\sqrt{\frac{\hbar^2}{2mL^2U}}$. Plot the two sides as functions of kL. Solutions are where the two curves intersect.

3. Apr 11, 2013

### amr55533

Thanks! I plugged the values in for "c" and ended up with the functions sin(kL) and 0.132(kL). I plotted both functions and found that they intersected at kL=0 and kL~0.8rad.