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Thermionic emission

  1. Jun 7, 2015 #1

    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
  2. jcsd
  3. Jun 8, 2015 #2

    Simon Bridge

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    Good questions - but they seem to be a matter of definition:
    1. check what K represents.
    2. check what "thermionic emission" means.
  4. Jun 8, 2015 #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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  5. Jun 8, 2015 #4


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    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.

  6. Jun 8, 2015 #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
  7. Jun 8, 2015 #6


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

  8. Jun 8, 2015 #7
    Thank you, I will read it as soon as I can.
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