How Far Must Caesium Atoms Be for a 1% Tunneling Probability?

[Rorshach]

Here is one problem I have, I think it is simple, but then again I thought that every problem I had was like that only to discover that there was some trick in the middle, so please help.

Caesium has exit work Wut = 1.9 eV. This means, therefore, that there are electrons in the cesium such that it is enough to give such an electron the extra energy 1.9 eV to make it free so that it can leave the surface. If two pieces of cesium are very near each other, such an electron can tunnel from one piece to the other and vice versa. What should the distance between the pieces be if the probability of tunneling will be 1%?
Line 1: use the approximate expression for the tunnelling.
Line 2: the electron has a specific energy E in the material. If so, what corresponds to the term V0 − E in this context?

Homework Equations

D=T*(T^conjugate)=\frac{4λ²μ²}{((λ²+μ²)²*sinh²(λa+4λ²μ²))}
λ=\frac{\sqrt{2m(U-E)}}{\hbar}
μ=\frac{\sqrt{2mE}}{\hbar}

The Attempt at a Solution

I tried to approach this as a problem of a finite potential barrier with three schrodinger equations, that would give the equation for transmission probability. They ask for approximated expression, so I think it is:
D=16*\frac{E}{u}*(1-\frac{E}{u})*exp(-2λa)

and a is the distance I am looking for.

 

[Rorshach]

anybody?

 

[TSny]

D=16*\frac{E}{u}*(1-\frac{E}{u})*exp(-2λa)

You do not have enough information to determine E and U separately. However, I have sometimes seen approximations where the factor in front of the exponential is taken to be of order 1 and the transmission probability is approximated by the exponential factor alone.
D \approx exp(-2λa)
See for example the middle of page 11 http://courses.physics.illinois.edu/phys485/web/tunneling.pdf .

If you do this, then you should be able to find $a$. You just need to think about how to use the value of the work function to determine the value of $\lambda$.

 

[Rorshach]

Maybe if I knew how to calculate energy for the electron on the last shell, I could use it as E, then U would be the the "height" of the potential wall, then it would be the exponential function with factor different than 1, but it wouldn't do much else. For the equation it would come down to ln(D)=2λa, where again I would need the knowledge of energy level on Caesium's last shell to go forward.

 

[TSny]

Note that λ only involves the difference between U and E. How is U-E related to the work function?

 

[Rorshach]

U-E is the work function, forgot about that, thanks:) If I was to do it the way You described (probability is rounded to exponeential factor only), it is enough to calculate a (which is more or less 3.91*10^24 m, though is looks a bit ridiculous to me, maybe there is some error on the way). However, if I wanted to calculate it using the original formula- I would need the energy of the electron on the last shell of caesium atom, so I could obtain those extra factors Tou suggested we substitute with 1.

 

[TSny]

As you noted, your result for $a$ is way off. I get a reasonable answer. You do need to be careful with the units.

Note that we are not dealing with individual Cs atoms here. Rather, we have solid Cs in which there are free electrons roaming around. So, the energy E is not the energy of the last shell of the atom, but rather the highest energy of the free electrons, which is related to the "Fermi Energy" . See http://en.wikipedia.org/wiki/File:Work.function.defined.gif .

 

[Rorshach]

okay, in the end I solved this by proposing two separate solutions: first one with the original formula
D=16∗Eu∗(1−Eu)∗exp(−2λa), where I used Fermi energy for caesium as U, and the result came out as 4.4 angstrem, and the second one, for which I used the simplified expression as suggested by TSny
D≈exp(-2λa)
in case of which the result came out to be 3.25 angstrem. I guess this concludes this problem, thank You very much for help:)