What Is the Wave Function of a Free Electron with 10 eV Kinetic Energy?

[Rorshach]

Hello guys, problem is as follows:
X9) A free electron has energy (kinetic) 10 eV and moves along the positive x-axis.
a) Determine the electron's wave function.
b) The electron reaches a potential step,-V0. Determine V0(expressed in eV), so that the probability of reflectance is 25%.
c) What is the reflection coefficient of the potential step instead is + V0 but with the same V0?

Homework Equations

ψ(x,t)=Cexp(i(kx-ωt))

The Attempt at a Solution

Equation above should do (with estimating the constant C) with the first paragraph, however I have problems with normalisation of this function, since there are no boundaries for the particle.

 

[TSny]

You won't need to determine a value of C. In addition to the incoming wave, you will have waves for reflection and transmission which also have constants. A certain ratio of the constants will determine the reflectance and you can determine the ratio without having to determine the constants themselves.

 

[Rorshach]

ok, I think I got it. Please tell me if am wrong:
for a) Psi(x,t)=Aexp(i(kx-(t*hbarred*k^2/2m)))
for b) V0=8*E
for c) reflection coefficient =0

am I right?

 

[TSny]

ok, I think I got it. Please tell me if am wrong:
for a) Psi(x,t)=Aexp(i(kx-(t*hbarred*k^2/2m)))
for b) V0=E
for c) reflection coefficient =0

am I right?

No, I don't think those answers are correct. You need to specify the full wave function in both regions (before and after the step). How did you get your answers for b) and c)?

 

[TSny]

ok, I think I got it. Please tell me if am wrong:
for a) Psi(x,t)=Aexp(i(kx-(t*hbarred*k^2/2m)))
for b) V0=8*E
for c) reflection coefficient =0

am I right?

I think your answer for b) is now correct.

I don't get the same answer as you for c)

 

[Rorshach]

answer I got for c) came from a rule I read in Griffiths introduction to Quantum Mechanics, that if E<V0, then T=0. But it only came from a rule, I assumed it as true and didn't push further.

 

[TSny]

If you square the h-bar in your answer for (a) then I think that would be correct if there is no step potential for this part.

I'm assuming that for part (b) the negative sign in -Vo means that the potential steps down as you move toward positive x.

 

[TSny]

answer I got for c) came from a rule I read in Griffiths introduction to Quantum Mechanics, that if E<V0, then T=0. But it only came from a rule, I assumed it as true and didn't push further.

T represents the transmission coefficient. You are looking for the reflection coefficient.

 

[Rorshach]

Also- specify the wave function in both regions- I thought it only concerned the situation when a prticle moves to the right(positive in x-axis), so I didn't have to include the step in estimating wave function. But if I was to do this- it would only mean that wave function would be divided into regions-
psi={
psi=Aexp(ikx)+Bexp(-ikx) for x<0
psi=Fexp(ilx) for x>0

 

[TSny]

Also- specify the wave function in both regions- I thought it only concerned the situation when a prticle moves to the right(positive in x-axis), so I didn't have to include the step in estimating wave function. But if I was to do this- it would only mean that wave function would be divided into regions-

psi=Aexp(ikx)+Bexp(-ikx) for x<0
psi=Fexp(ilx) for x>0

That looks good.

 

[Rorshach]

ok, since transmission coefficient is equal to 0, and transmission coefficient and reflection coefficient summed up have to be equal to 1, then reflection coefficient is equal to 1, am I right?

Also, I tried to derivate that formula for free particle, but still had normal non squared hbarred, but will try to derivate again.

 

[TSny]

ok, since transmission coefficient is equal to 0, and transmission coefficient and reflection coefficient summed up have to be equal to 1, then reflection coefficient is equal to 1, am I right?

That's correct.

Also, I tried to derivate that formula for free particle, but still had normal non squared hbarred, but will try to derivate again.

Oops, my mistake. The energy has $\hbar^2$, but the wavefunction will not have the $\hbar$ squared. You were correct. Sorry.