# Homework Help: Potential barrier problem

1. Sep 10, 2010

### zak8000

1. The problem statement, all variables and given/known data
a particle of kinetic energy E is incident from left on a potential barrier,height U, situated at the origin.the barrier is infinitely wide and E>U

obtain an expression for the reflection coefficient R of the particle as a fuction ratio e=E/U

2. Relevant equations

3. The attempt at a solution

to left of barrier wavefunctions are free particle waves
barrier at x=0
psi(x,t)=Aexp(ikx-wt)+ Bexp(-ikx-wt) x<0

within barrier wavefunction also is oscillatory

E=h(cross)*w

considering case for E<U and using TISE on psi(x,t) within barrier we get

a=([2m(U-E)]^0.5)/h(cross)

but now E>U and as a result a becomes imaginary. introducing new wavenumber L and barrier wavefunction becomes

psi(x,t) =Cexp(-iLx-wt)+Dexp(iLx-wt) x>0 (note this is equation is only for a barrier of finite width)

but now everywhere to right from origin x=0 is the barrier wavefunction given above . to keep psi(x,t) from diverging for large x we must take D=0 leaving only decaying wave and this is where i am up to i was just wanting to know if i am on the right track

2. Sep 11, 2010

### kuruman

You need two sets of wavefunctions, one for x<0 and another one for x>0. The first set for x<0 has two pieces, one representing the incident wave and one representing the reflected wave. For x>0 you have only one wave, the transmitted wave traveling to the right. Since E > U, all waves are represented by complex exponentials. These are basically sinusoidals and do not decay with x.

3. Sep 11, 2010

### zak8000

ok i see thanks so now my wavefunctions are

psi(x,t)=Aexp(ikx-wt)+Bexp(-ikx-wt) x<0
psi(x,t)=Dexp(iLx-wt) x>0

so now wavefunctions must be joined smoothly following the conditions
A+B=D cont of psi
ikA-ikB=iLD cont of d(psi)/dx

solving for D i get

A(1-k/L)=B(-1-k/L)

B/A= -(1-k/L)/(1+k/L)

reflection coefficient is given by R=|B^2|/|A^2| but i need to obtain an expression in terms of ratio E/U

so i tried to substitute k=(2mE/h(cross))^0.5 and L=i(2m(U-E)/h(cross))^0.5 into equation but was unsuccessful to get the ratio out am i on the right track?

4. Sep 11, 2010

### kuruman

Aren't k and L both real and isn't the ratio L/k related to U/E somehow?

5. Sep 11, 2010

### zak8000

yes sorry both L and k are real and i get E/U =-1/2 when i put them equal to each other but now im kind of lost

6. Sep 12, 2010

### kuruman

I don't see how you get E/U = -1/2. What are your (correct) expressions for k and L?