# Probability that the electron will pass through the barrier?

1. Sep 27, 2013

### Gee Wiz

1. The problem statement, all variables and given/known data
An electron of energy E = 2.2 eV is incident on a barrier of width L = 0.7 nm and height Vo = 3.2 eV as shown in the figure below. (The figure is not drawn to scale.)

What is the probability that the electron will pass through the barrier?
The transmission probability is?

2. Relevant equations

T=Ge^(-2kL)
G=16(e/v)(1-e/v)
k=sqrt(2m/(h/2∏)^2(U-E)
3. The attempt at a solution
So I thought this was straight forward and was putting all the values into the formulas and I keep getting 0. Which is not correct...is there something I am overlooking?

2. Sep 27, 2013

### Gee Wiz

I got G = 3.4375 eV
k= 5.125e9 J

Am I missing that T is just a coefficient and I should use it as something else..?

Last edited: Sep 27, 2013
3. Sep 27, 2013

### vela

Staff Emeritus
The units you have for G are wrong, though the number is right. The units of $k$ should be 1/length, and it's way off numerically.

It's useful to memorize certain combinations of constants:
\begin{align*}
m_e c^2 &= 511000\text{ eV} \\
\hbar c &= 197\text{ eV nm}
\end{align*} where $c$ is the speed of light. Your expression for $k$ can be rewritten slightly by introducing convenient factors of $c$:
$$k = \sqrt{\frac{2m_e(U-E)}{\hbar^2}} = \sqrt{\frac{2(m_ec^2)(U-E)}{(\hbar c)^2}}$$

Last edited: Sep 27, 2013
4. Sep 27, 2013

### Gee Wiz

Okay, that makes sense so for k I get 5.131. So then I took
T=3.4375e^(-2*5.131*.7)

5. Sep 28, 2013

### Gee Wiz

I think I am still missing something, because I keep getting .002607 at my T, yet this is not correct

Last edited: Sep 28, 2013
6. Sep 28, 2013

### rude man

here's a nifty site for this. Their formula is correct, I think it's the same as yours.

HOWEVER: they also have a calculator and when I inputed your parameters I get that the probability is 1.0!

I don't know if it's right or wrong, but there it is.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html

7. Sep 28, 2013

### vela

Staff Emeritus
How do you know it's not correct?

8. Sep 28, 2013

### Gee Wiz

I tried that, but unfortunately it was not correct. But thank you for the suggestion.

9. Sep 28, 2013

### Gee Wiz

I know it is not correct because the online homework marks it incorrectly...but maybe the system is having issues..?

10. Sep 28, 2013

### vela

Staff Emeritus
Are you sure you're using the right formulas? i get the same result you do, but I just assumed you looked up the right formulas. It seems unlikely that hyperphysics would have an erroneous page up.

It could also be you're entering the answer incorrectly, e.g. sig figs, units, etc.