# Thermionic emission

1. Jun 7, 2015

### chimay

Hi,

in the latest lesson my professor explained the thermionic emission; I guess it is a simplified approach, since I am not a Physicist.
Anyway, there are some things not clear to me; I'll show you the approach and I hope someone of you could help me.

As usual, $W=$ work function and $E_{f}=$ Fermi Level and $J=$ current density
The solution is straightforward:
$$J= -q \int 2 \frac{d^{3} \bar k\ }{(2\pi)^{3}} \frac{1}{e^{\frac{E_{z}+E_{||}-E_{f}}{kT}+1}} \frac{1}{\hbar} \frac{dE_{z}}{dK_{z}}$$
$E=\frac{{\hbar}^{2}}{2m^{*}} ( {K_{x}}^{2}+{K_{y}}^{2}+{K_{z}}^{2} )= E_{||}+E_{z}$ and $z$ is the direction of flowing of the current.

$2 \frac{d^{3} \bar k\ }{(2\pi)^{3}}$ is the density of states per unit volume

$\frac{1}{e^{\frac{E_{z}+E_{||}-E_{f}}{kT}+1}}$ is the probability of occupation of each state

$\frac{1}{\hbar} \frac{dE_{z}}{dK_{z}}$ is the velocity of each carrier

Since $d^{3} \bar k\ = d^{2}k_{||}*dK_{z}$
$$J= \frac{-2q}{{2\pi}^{3}\hbar} \int d^{2}k_{||} \int_{E_0}^{+\infty} dE_{z} \frac{1}{e^{\frac{E_{z}+E_{||}-E_{f}}{kT}+1}}$$
$E_{0}= E_{f}+W$

My questions are the following ones:
- the integral goes from $E_{0}$ to $+ \infty$; since electrons that have energy higher than $E_{0}$ are "free", why are concepts like the K-space still used? Are they related to electrons bounded in a crystal, are they?
- in the integral there is no reference to any electric field applied. I can't understand how can we have a net current just by summing thermal velocities.

Thank you

2. Jun 8, 2015

### Simon Bridge

Good questions - but they seem to be a matter of definition:
1. check what K represents.
2. check what "thermionic emission" means.

3. Jun 8, 2015

### chimay

Your answer is not clear to me , should it be a kind of "hint" to solve my questions?

I have thought about the problem since yesterday, but I really can't solve it.
Suppose for a moment that an electron with energy above $E_{0}$ can still be described by means of K-Space (I am not convinced about this, it is an assumption).
Basically, the explanation of the integral is the following: let's choose a surface $S$, for each $K$ we have a certain current density $J= -qn(K)V_{z}(K)$, let's sum them all togheter and we get the overall current. It seem like taking a piece of metal and expecting a current flowing just by thermal agitation.

Obviously this is no the case of thermion emission, since the elctrons escaping from surface (please refer to the figure) do not have a counterpart coming inside the metal, my problem is how this is translated into mathematical language.

Thank you very much

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4. Jun 8, 2015

### ZapperZ

Staff Emeritus
No. Every electrons will have a momentum. Even in the free-electron gas model, you have a distribution of momentum. So that is why the description is in k-space. In fact, in solid state physics, this is the most useful representation, because especially in any diffraction experiments, the results are equivalent to probing the k-space structure.

Again, there is a distribution of momentum of the electrons. There will be a net flow of electrons coming out of the solid due to the fact that they have a momentum component perpendicular to the surface. So you will get a current.

BTW, a more detailed explanation on this is via the Richardson-Dushman model.

Zz.

5. Jun 8, 2015

### chimay

Actually, We have not studied the free-electron gas model.
Could you please give an explanation (even qualitative) without referring to it?

Thank you

6. Jun 8, 2015

### ZapperZ

Staff Emeritus
http://www.inorg.chem.uni-sofia.bg/Courses/03/Freelgas.pdf

In particular, look at the dispersion relation for the free-electron gas.

Zz.

7. Jun 8, 2015

### chimay

Thank you, I will read it as soon as I can.