PS # 61 from 4R

[Apparition]

An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected?

A) twice as large
B) approximately equal
C) 1/4 as large
D) 0

The answer is that it will be equal. What I don't understand is doesn't the positive anode exert an attractive electrostatic force on the
electron causing it to accelerate?

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[willthatsall]

Pretty sure the anode is positive in this problem. Since the energy of the photon is basically the same as the work function, the electron is ejected with ~0 kinetic energy. So the final kinetic energy is equal to the initial potential. Remember that anodes are not always negative, it depends if the cell is galvanic or electrolytic.

 

[Apparition]

Pretty sure the anode is positive in this problem. Since the energy of the photon is basically the same as the work function, the electron is ejected with ~0 kinetic energy. So the final kinetic energy is equal to the initial potential. Remember that anodes are not always negative, it depends if the cell is galvanic or electrolytic.

Yeah, I reread the passage and realized that it's positive. So I changed it in the question. So basically it's just a conversion of PE to KE. Thanks!!

 

[JDAD]

yeah, it is B, when it is ejected, its kinetic energy is just enough to get escape, most of it will go to overcoming the work function. Therefore, you assume it has almost no kinetic energy.

It at the moment has a potential energy and no kinetic energy, when it reaches the anode, all of its potential energy has been converted to kinetic energy. Therefore, the intitial potential energy is almost equal to its final kinetic energy.

The charge on the anode doesn't matter. In this case it is actually positive. You are told that the electron will go to the anode, so think of it like a mass in the air. It has a potential energy, and very little kinetic energy.

When I took that test I misread the question and screwed it up. I read potential as kinetic and that is why I picked A, the kinetic energy final will be greater than intial, but that isnt' what the questions asked for. So be careful when you read the question.

 

[JDAD]

wow, while i was typing that out you already got your answer. i feel like a tool

 

[Apparition]

wow, while i was typing that out you already got your answer. i feel like a tool

Thanks for your help! I actually picked 0 because I thought it would stick to the anode and that's it lol. Most of the mistakes come from misreading something, it's annoying.

 

[Csv321]

I figured since I just finished the physical science section and tripped on that passage a little bit, I'd post what the solutions manual says...it's pretty much the same as what everyone else has already said

Oh...and I just barely made a 9 on PS!!!

From 4R Solutions:

Since the electron is ejected by a photon whose energy is only slightly greater than the work function of the metal, there is very little energy left over as kinetic energy of the electron. So, all of the photoelectron's energy after its release from the cathode is its electrical potential energy. This energy is then converted to kinetic energy as the electron accelerates across the gap to the anode.