- #1
skrat
- 748
- 8
Homework Statement
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?
Homework Equations
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?