Thermionic Emission: Questions Explained

In summary, my professor explained the thermionic emission. There are some concepts that are not clear to me, but I hope someone of you could help me. There are some questions that I still haven't been able to answer.
  • #1
chimay
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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, [itex]W=[/itex] work function and [itex] E_{f}= [/itex] Fermi Level and [itex] J= [/itex] current density
The solution is straightforward:
[tex] 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}} [/tex]
[itex] E=\frac{{\hbar}^{2}}{2m^{*}} ( {K_{x}}^{2}+{K_{y}}^{2}+{K_{z}}^{2} )= E_{||}+E_{z}[/itex] and [itex]z[/itex] is the direction of flowing of the current.

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

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

[itex]\frac{1}{\hbar} \frac{dE_{z}}{dK_{z}} [/itex] is the velocity of each carrier

Since [itex] d^{3} \bar k\ = d^{2}k_{||}*dK_{z} [/itex]
[tex] 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}}[/tex]
[itex] E_{0}= E_{f}+W [/itex]

My questions are the following ones:
- the integral goes from [itex] E_{0} [/itex] to [itex] + \infty [/itex]; since electrons that have energy higher than [itex] E_{0} [/itex] 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
 
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  • #2
Good questions - but they seem to be a matter of definition:
1. check what K represents.
2. check what "thermionic emission" means.
 
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  • #3
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 [itex] E_{0} [/itex] 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 [itex] S [/itex], for each [itex] K [/itex] we have a certain current density [itex] J= -qn(K)V_{z}(K) [/itex], 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
chimay said:
My questions are the following ones:
- the integral goes from [itex] E_{0} [/itex] to [itex] + \infty [/itex]; since electrons that have energy higher than [itex] E_{0} [/itex] are "free", why are concepts like the K-space still used? Are they related to electrons bounded in a crystal, are they?

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.

- 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

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.
 
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  • #5
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
chimay said:
Actually, We have not studied the free-electron gas model.
Could you please give an explanation (even qualitative) without referring to it?

Thank you

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

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

Zz.
 
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  • #7
Thank you, I will read it as soon as I can.
 

FAQ: Thermionic Emission: Questions Explained

What is thermionic emission?

Thermionic emission is the process by which electrons are emitted from a heated material, typically a metal, into a vacuum or low pressure gas environment.

How does thermionic emission work?

In thermionic emission, electrons gain enough energy from heat to overcome the attractive forces of the metal and are emitted into the surrounding space. This creates a flow of electrons, known as an electron current.

What is the significance of thermionic emission?

Thermionic emission plays a crucial role in many electronic devices, including vacuum tubes, cathode ray tubes, and electron microscopes. It also has applications in power generation and spacecraft propulsion.

What factors affect thermionic emission?

The rate of thermionic emission is affected by the temperature of the material, the work function of the material, and the electric field near the material's surface.

How is thermionic emission related to the photoelectric effect?

Both processes involve the emission of electrons from a material, but thermionic emission is caused by heat rather than light. However, both processes follow the same basic principles of electron emission and can be described by the same equations.

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