Flow of electrons hit a potential hole

1. Jan 11, 2014

skrat

1. The problem statement, all variables and given/known data
Flow of 500 electrons per second with kinetic energy 3 eV hits a perpendicular 5 eV potential hole 0.3 nm wide. How many electrons pass per second pass the obstacle?

2. Relevant equations

3. The attempt at a solution

Hmm, I checked my notes where it is written that coefficient of electrons that passes the obstacle is calculated as $T=(1+\frac{1}{4}(\frac{k_1}{\kappa }+\frac{\kappa }{k_1})^2sinh^2(\kappa a))^{-1}$

Where I used notation $k_1=\sqrt{\frac{2mE}{(h)^2}}$ and $\kappa =\sqrt{\frac{2m(V-E)}{(h)^2}}$. I don't know how to write crossed h in latex, so I used (h) instead. Notation a tells how wide the hole is.

So $k_1=8.66 nm^{-1}$ and $\kappa =7.07 nm^{-1}$ and $sinh^2(\kappa a)=0.00137$

Which gives me $T=0.05377$ and therefore 26 electrons should pass the obstacle. BUT the result in the book states 408 electrons as result...

Doesn anybody know what am I doing wrong here?

2. Jan 11, 2014

TSny

Not sure what potential "hole" means. But if it means a potential "well" of depth 5 eV, then the kinetic energy will still be positive inside the well. So, you will have an oscillatory wavefunction inside the well rather than exponential behavior. Instead of a "kappa" $\kappa$, you'll have a $k_2$ wavevector inside the well. What happens to the Sinh function in this case?

3. Jan 12, 2014

skrat

Potential well it is. In direct translation from my language it is a hole. :)

Now here is my question. How will the kinetic energy still be positive inside the well? Before the well it is 3 eV and the well has a depth of 5 eV.

so $k_2 =\sqrt{\frac{2m(E-V)}{(h)^2}}=i\sqrt{\frac{2m(v-E)}{(h)^2}}=i\kappa$

In case you are right, which you probably are but I would like to understand why... sinh is than sin function.

4. Jan 12, 2014

TSny

If you take the potential to be 0 outside the well, then inside the well it will be -5eV. The kinetic energy is the difference between E and V: E-V. This gives a positive value of the KE inside the well.

Right, the sinh function becomes a sin function.

5. Jan 12, 2014

skrat

So E-V=8 eV.

Than, only if V is positive than wavevector $k_2$ will be complex, or... ?

6. Jan 12, 2014

TSny

If V is positive and greater than E, then the wavevector will be imaginary. So, if you had a potential barrier of height 5 eV with E = 3 ev, then the kinetic energy E-V would be negative inside the barrier.

7. Jan 12, 2014

skrat

How does this differ from my original post (problem)?

8. Jan 12, 2014

TSny

The original post dealt with a potential well (I think), whereas, my last comment was for a potential barrier.

9. Jan 12, 2014

skrat

For a moment a thought that's the same.