- #1
chimay
- 80
- 6
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
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
