How does thermionic emission work?

  • #1
I’ve been curious about understanding the mechanism behind themionic emission from what I have read I found that themionic emission happen when the energy from added temperature excess the work function of the material. I also readed when temperature excesses 1000k themionic emission happens but when I calculate the energy in Ev from that temperature I get a number no where close to the work function of tungsten a material commonly used as filament for electron guns so what gives?
Wien Law
~~~~~~~~~~~2.89e-3
Wavelength =————-
~~~~~~~~~~~~1000k
Wavelength = 289e-8m

Conversation of wavelength to Joules
~~(6.626e-27)(3e10)
E=—————————
~~~~~~289e-8m
E=0.06878e-18
Joules to Electron volts
E = 0.06878e-18 * 6.242e18

E = 0.4293ev
Tungsten Workfunction = 4.5ev
I not sure if I’m just calculating the temperature with the wrong equation but those numbers don’t seem anywhere closes , when I used the work function of tungsten to calculate the temperature I get a number that is insanely high

Ev to wavelength
~~~~~~~~~~~(6.626e-34)(3e10)
Wavelength= —————————-
~~~~~~~~~~~~(4.5ev)/(6.242e18)

Wavelength = 2.757e-7
Wavelength = 275.7nm

Wien law
~~~~~~2.89e-3
T(k) = ————
~~~~~~2.757e-7
T(k) = 10482.408k

10482 KELVIN!!! that way excess the melting temperature of tungsten and that far exceeds the temperature in any electron gun if I am calculating the temperature wrong how do I do it properly and can anyone confirm that thermionic emission happens from temperature exceeding work function?

I read that thermionic emission happens from exceeding the work function of material from it Wikipedia page(I know it’s not the most credible site)

I also read that the temperature for thermionic emission of t > 1000k from Wikipedia also.
Sorry for any grammar mistakes I’m not a very good at spotting them
 

Answers and Replies

  • #2
Charles Link
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In the Wien's law calculation for the peak wavelength of the Planck blackbody emission spectrum, ## 4.96=\frac{E_{\lambda \, max}}{k_B T} ##, so that ## E_{\lambda \, max}=4.96 \, k_B T ## gets you a better number than if you just use ## k_B T ##, which is ## 1.38 \cdot 10^{-20}/1.602 \cdot 10^{-19} =.086 \, eV ## for ## T=1000 ##, and compare it to the work function ## W ##. Part of the answer would seem to be in the ## T^2 ## factor in the Richardson-Dushman equation. See http://simion.com/definition/richardson_dushman.html
 
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  • #3
sophiecentaur
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Thermionic cathodes do not use Tungsten as a source of electrons. They are coated with a range of materials. See this wiki link about hot cathodes. The operating temperature of many valve cathodes is less than 600°C (also mentioned in the link).
Edit: What have you read up about thermionic emission? Electrons do not totally escape from a hot cathode if there is no handy positive electrode nearby. Without an external field, they hang around the surface and return to the cathode which has become positively charged. Electrons do not need 'escape velocity'.
 
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