- #1
chimay
- 81
- 8
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
