Work function and energy of X ray photon emitted by anode

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SUMMARY

The discussion centers on the energy of X-ray photons emitted by anodes in relation to the work function of electrons. Participants clarify that the energy of the emitted photon is represented as E=hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon. The work function is considered negligible in this context, as the initial kinetic energy of the electron is primarily derived from thermal energy and the accelerating voltage (eV). The consensus is that the work function does not directly factor into the energy of the emitted photon when considering the energy transfer during the collision with the anode.

PREREQUISITES
  • Understanding of Planck's equation (E=hf)
  • Knowledge of the photoelectric effect and work function
  • Familiarity with electron acceleration in cathode-anode systems
  • Basic principles of X-ray production and photon emission
NEXT STEPS
  • Research the relationship between photon energy and electron kinetic energy in X-ray tubes
  • Study the implications of work function in different materials, particularly anodes
  • Explore the photoelectric effect in detail, focusing on energy transfer mechanisms
  • Investigate the role of thermal energy in electron emission from cathodes
USEFUL FOR

Physicists, electrical engineers, and students studying quantum mechanics or X-ray technology will benefit from this discussion, particularly those interested in the principles of photon emission and electron dynamics in cathode-anode systems.

SecretSnow
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Hi guys, I'm constantly bothered by one assumption in my textbook..it says that the photon emitted by the accelerating electron boiled off from the cathode colliding into the anode, has E=hf neglecting the work function, since its negligible. I'm curious whether it's E=hf plus or minus the work function, and I'm really confused by this. Firstly, if I work backwards, like imagining it's a photon liberating the electron through the photoelectric effect instead, then the energy released should be E=hf + work function right? Since it emits the work function energy of the electron together with the photon right? However, how do we know they are separate values, since the work function energy can itself be incorporated into the hf value of the photon? Wouldn't it be the same then?

Next, I think it's E=hf - the work function since if the accelerated electron has an initiall KE of E=eV=hf-work function then wouldn't the emitted photon have energy E=hf -phi too? Hmm but I think this whole second paragraph is invalid since the electron is initially boiled off by heating right? I'm not sure...I hope you guys can help me clear my confusion! Although the reason is not really needed, I'm curious to find out, thanks a lot!
 
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The initial electron cannot have a negative energy at the cathode - it starts with at least zero (and the required energy to exit the material is thermal energy), and has at least eV at the time it hits the anode. If the electron directly gets captured in the anode, hf=E=eV+work function might be possible as photon energy.
 
Actually I don't get why the textbook even mentions work function in the first place. if the electron is boiled off,then initially if should have low or zero KE, which means by the time it reaches the anode, it should have E=eV. The thing is, does the work function applies in the anode? If yes, then I'm assuming after it reaches the anode it goes into the ground state right? (Since it's been decelerated to the lowest energy state possible) during this acceleration, the photon should have E=hf, now, I think, without work function, or it is inside the E=hf equation already. Where then does the work function even apply? Thanks a lot!
 
I'm thinking if it could be the work function of anode it's saying, and that there might be electrons of the anode absorbing the photon produced by the colliding electrons? Is this possible? But if the energy of the photon emitted is high enough, then it should be able to liberate the electrons on the anode with kinetic energy supplied too, so it isn't insignificant is it? It's more like assuming the pathway of the photon doesn't collide with the anode...
 

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