How can I find the maximum KE of the photoelectrons?

[sodnaz]

Homework Statement

I don't know how to do Question 2 part C.

Homework Equations

I know that E=hf and I know the photoelectric equation: hf = work function + 1/2mv^2

The Attempt at a Solution

I honestly don't have a clue how to do part c of question 2. I think it involves one of the two equations I said above, but I'm not sure which one it would be and why. Could someone help me please?

Thank you

 

[kuruman]

You are looking for kinetic energy, so which of the equations you posted looks more promising?

 

[sodnaz]

You are looking for kinetic energy, so which of the equations you posted looks more promising?

I'm not sure, since they would both be kinetic energy wouldn't they (since for E=hf, the photons have kinetic energy since they are moving)? Or is the first one (E=hf) not for kinetic energy?

 

[kuruman]

Photons have energy and momentum but zero mass. Their energy is E = hf and their momentum is p = hf/c. Besides, you are asked to find the kinetic energy of the electrons and that's KE = ½mv².

 

[sodnaz]

Photons have energy and momentum but zero mass. Their energy is E = hf and their momentum is p = hf/c. Besides, you are asked to find the kinetic energy of the electrons and that's KE = ½mv².

But since the photons are traveling through the air, their energy must be kinetic energy, right? I know I need to find the KE of the photoelectrons but I'm asking about normal photon energy here, the incident photons.

 

[kuruman]

But since the photons are traveling through the air, their energy must be kinetic energy, right?

Photons have only one kind of energy and that's hf. All photons have zero mass and travel at the speed of light in vacuum. Thus, it doesn't make sense to talk about their kinetic energy ½mv². That's why we do not attach the qualifier "kinetic" to the photon energy.

In any case, do you understand what the photoelectric equation says? Can you explain it in simple terms? Understanding that should help you figure out how to answer the question.

 

[sodnaz]

Photons have only one kind of energy and that's hf. All photons have zero mass and travel at the speed of light in vacuum. Thus, it doesn't make sense to talk about their kinetic energy ½mv². That's why we do not attach the qualifier "kinetic" to the photon energy.

In any case, do you understand what the photoelectric equation says? Can you explain it in simple terms? Understanding that should help you figure out how to answer the question.

Ah, that explains the question regarding the KE, thank you.

I know that the photoelectric equation basically says, a photon of energy hf has to overcome the work function in order to release a photoelectron, and any "spare energy" left over (given by KE=hf-work function) is KE of the photoelectron.

I don't know how I use that equation with part c though. Do I just say that the KE = hf - work function? If it's that easy, how would you integrate it with those energy levels shown just below the question? How do they come into these equations?

 

[kuruman]

Do I just say that the KE = hf - work function?

You just say that. Think of the electron as absorbing all of the photon's energy. It uses some of the energy (equal to the work function) to pay the metal for its freedom. Any energy that's left after the electron leaves the surface, is kinetic energy that it spends however the circumstances dictate.

If it's that easy, how would you integrate it with those energy levels shown just below the question? How do they come into these equations?

It's that easy, but unrelated to the energy levels. That part of the question has to do with photon (not electron) emission by a single atom. The energy of the emitted photon must match the energy difference between levels.

 

[sodnaz]

You just say that. Think of the electron as absorbing all of the photon's energy. It uses some of the energy (equal to the work function) to pay the metal for its freedom. Any energy that's left after the electron leaves the surface, is kinetic energy that it spends however the circumstances dictate.

It's that easy, but unrelated to the energy levels. That part of the question has to do with photon (not electron) emission by a single atom. The energy of the emitted photon must match the energy difference between levels.

For question 3a: So if the energy of the emitted photon matches the energy differnece, and change in E = hf, then to find the frequency it would be change in energy/planck's constant, aka delta E/h = f? Then I use that value of f in E=hf?

 

[kuruman]

... then to find the frequency it would be change in energy/planck's constant, aka delta E/h = f?

Yes.

Then I use that value of f in E=hf?

No reason to do that. You already know that the energy of the photon is the energy difference between levels. You will be going around a circle if you did that.