Doubt regarding conversion of eV to V

[Gh0stS3C]

Homework Statement

The ejection of the photo electron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Relevant Equations

Equation used for calculating Work Function :
Energy of photon = Work function + Kinetic Energy {EQ 1}

Work Function = Energy of photon - Kinetic Energy {From EQ 1}
---------------------------------------------------------------------------------------------------------------------------
Energy of photon = hc/lambda [h = Planck's Constant, c = speed of light, lambda = wavelength]

So I know exactly how to solve this question with the relevant equations. However problem I ran into was how I substitute for K.E (Kinetic Energy). I asked one of my teachers for help. She said that since the voltage applied stops the experiment, it is equivalent to the energy of the electron which she then proceeded to take as 0.35 eV (This is what I made out from her explanation, studying through online classes is proving to be very difficult for me.) Can somebody walk me through the process of how we arrived at 0.35 eV as the K.E.

My attempt so far :

Wavelength = 256.7 nm
converting to metres --> 256.7 x 10^-9 m

Energy of photon = 6.626 x 10^-34 x 3 x 10^8/256.7x10^-9

I cannot proceed further due to my doubt regarding eV conversion but I know what to do after I get the K.E value in eV.P.S Please let me know of any tips to make my question more clear. I'm new here but looking forward to learn :)

 

[etotheipi]

You eject the electron and the electric field exerts a force to slow it down just before it reaches the other plate. The work done by the electric field on the electron if the change in potential $\Delta V$ is $-0.35V$ is $W = -\Delta U = -q\Delta V = e\Delta V = (-0.35e) \ \text{J}$.

What does the work energy theorem state?

 

[BvU]

Hello spooky, !

Keep going ! You are doing well.

But don't forget units ! What result do you get for the energy of the photon and in what units is that ?

Do you know how to convert that to the units in which work functions are usually expressed ? E.g. in this table ?

About the photoelectric effect: It is known that photons with enough energy can knock electrons out of a metal. Enough energy means "enough to overcome the work function". Any excess can be used to provide kinetic energy to the liberated electron. By applying a blocking voltage (in your exercise apparently a minimum of 0.35 V) the electrons can be prohibited from reaching the other electrode.

 

[Gh0stS3C]

Hello spooky, !

Keep going ! You are doing well.

But don't forget units ! What result do you get for the energy of the photon and in what units is that ?

Do you know how to convert that to the units in which work functions are usually expressed ? E.g. in this table ?

About the photoelectric effect: It is known that photons with enough energy can knock electrons out of a metal. Enough energy means "enough to overcome the work function". Any excess can be used to provide kinetic energy to the liberated electron. By applying a blocking voltage (in your exercise apparently a minimum of 0.35 V) the electrons can be prohibited from reaching the other electrode.

Hi,
My bad on that forgot to mention that it was originally in Joules. After I read what you wrote I just realized that converting the Joules to eV should make the calculation more easier and also never knew that this was the unit usually used for work functions.

so an additional step here is divide the energy of photon by 1.602 x 10^-19. then we will obtain 4.83 eV.

I haven't understood yet how to apply the information on blocking voltage though. Like thinking question-wise I'm guessing I should be able to get a value for K.E from it however I haven't figured out how I do that

 

[etotheipi]

You attach the positive pole of a voltage source to the metal in question and the negative pole to a collector plate, so that the resulting electric field points from the metal to the collector.

When an electron is emitted from the metal, it still has some kinetic energy. We assume it is emitted normally to the surface of the metal so that it moves along a straight line between the two plates.

The idea is that electrons stop reaching the collector when the voltage across the plates is high enough that the work done by the electric field on the electron is negative enough to reduce its kinetic energy to zero before it reaches the collector. You know the amount of work done (post #2), and if you have the work-energy theorem at your disposal you should get the desired result.

 

[Gh0stS3C]

You eject the electron and the electric field exerts a force to slow it down just before it reaches the other plate. The work done by the electric field on the electron if the change in potential $\Delta V$ is $-0.35V$ is $W = -\Delta U = -q\Delta V = e\Delta V = (-0.35e) \ \text{J}$.

What does the work energy theorem state?

I haven't learned about the work energy theorem yet but from a quick Google search I think it means all forces applied on a body is transformed into kinetic energy (correct me if I'm wrong) but then I couldn't completely understand the last equation you wrote.

 

[etotheipi]

Work is the quantity you get if you multiply the component of the force in the direction of the displacement with the displacement of the particle.

The key part here is that the total work done by all forces (here there is only one: the electric force) equals the change in kinetic energy. And the change in kinetic energy here will be "zero minus the initial kinetic energy".

 

[Gh0stS3C]

Work is the quantity you get if you multiply the component of the force in the direction of the displacement with the displacement of the particle.

The key part here is that the total work done by all forces (here there is only one: the electric force) equals the change in kinetic energy. And the change in kinetic energy here will be "zero minus the initial kinetic energy".

I read through this carefully and I have a few questions.

1)Potential of 0.35 V stopped the photo electric effect from taking place so then this must imply that K.E was changed to zero right? (since it stopped the movement from taking place)
2) Based on what you said I am guessing the work done is equivalent to the value of the initial K.E. (since at the end K.E turns out to be 0 and for this to occur the change must be equivalent to the value of initial K.E)

Hopefully I'm right till here

 

[etotheipi]

1)Potential of 0.35 V stopped the photo electric effect from taking place so then this must imply that K.E was changed to zero right? (since it stopped the movement from taking place)
2) Based on what you said I am guessing the work done is equivalent to the value of the initial K.E. (since at the end K.E turns out to be 0 and for this to occur the change must be equivalent to the value of initial K.E)

Pretty much; although the change in kinetic energy is negative (i.e. it's moving and then it's not) and consequently the work done is also negative (which makes sense, since the electric force is in the opposite direction to the velocity i.e. slowing it down).

But you can correctly say the absolute value of the work done equals the absolute value of the change in kinetic energy, yes, which is basically equivalent algebraically.

 

[BvU]

so an additional step here is divide the energy of photon by 1.602 x 10^-19. then we will obtain 4.83 eV.

Good !

I haven't understood yet how to apply the information on blocking voltage though.

Well the experiment tells us that 0.35 V is just enough to prevent electrons from migrating to the other electrode. So apparently, with 4.83 eV photons you can kick out electrons with a maximum energy of 0.35 eV.

Like thinking question-wise I'm guessing I should be able to get a value for K.E from it however I haven't figured out how I do that

Just a small step now: how much of this 4.83 eV is needed to do the work of kicking it out of the metal ?

Compare with the value in the table.

An old cynic like me would say: Too good to be true . But then again, it's an exercise composed with the aim to get you on your way in this fascinating branch of science.

 

[Gh0stS3C]

Pretty much; although the change in kinetic energy is negative (i.e. it's moving and then it's not) and consequently the work done is also negative (which makes sense, since the electric force is in the opposite direction to the velocity i.e. slowing it down).

But you can correctly say the absolute value of the work done equals the absolute value of the change in kinetic energy, yes, which is basically equivalent algebraically.

Ah yes. So correcting myself : absolute value of work done is equal to absolute value of initial K.E of electron

so then value of K.E of the particle is 0.35. But which unit is this in? (Sorry if this question seems silly)

 

[etotheipi]

Well, the work done (I almost wrote 'worque', been doing too much rotational dynamics of late...) we found to be $W = e\Delta V = -0.35e \ \text{J}$.

Now, how many $eV$ is $e$ joules?

 

[Gh0stS3C]

Good !
Well the experiment tells us that 0.35 V is just enough to prevent electrons from migrating to the other electrode. So apparently, with 4.83 eV photons you can kick out electrons with a maximum energy of 0.35 eV.
Just a small step now: how much of this 4.83 eV is needed to do the work of kicking it out of the metal ?

Compare with the value in the table.

An old cynic like me would say: Too good to be true . But then again, it's an exercise composed with the aim to get you on your way in this fascinating branch of science.

Using the equation for threshold energy and the info we have now, I think it is 4.83 - 0.35. This means the answer i.e Work function of silver metal = 4.48eV
I'm pretty sure this is the answer right? :D

 

[etotheipi]

Using the equation for threshold energy and the info we have now, I think it is 4.83 - 0.35. This means the answer i.e Work function of silver metal = 4.48eV
I'm pretty sure this is the answer right? :D

Believe so, yes.

 

[Gh0stS3C]

Well, the work done (I almost wrote 'worque', been doing too much rotational dynamics of late...) we found to be $W = e\Delta V = -0.35e \ \text{J}$.

Now, how many $eV$ is $e$ joules?

Thanks so much sir! The feeling when you finally understand a concept is really something :D
this will be 1.602 x 10^-19 J which is equivalent to 1 eV! Amazing! My teachers like to skip a lot of the syllabus especially now with online class which leaves us confused a lot of the times. Thanks for the help guys!

 

[etotheipi]

Thanks so much sir! The feeling when you finally understand a concept is really something :D
this will be 1.602 x 10^-19 J which is equivalent to 1 eV! Amazing! My teachers like to skip a lot of the syllabus especially now with online class which leaves us confused a lot of the times. Thanks for the help guys!

Yeah, the electron volt is a slightly peculiar unit at first. I just remember that $(e) \text{J} \equiv (1) eV$.

It's a good idea also to have a read up about that work energy theorem we used, it pops up quite a lot and will be quite helpful!

 

[Gh0stS3C]

Yeah, the electron volt is a slightly peculiar unit at first. I just remember that $(e) \text{J} \equiv (1) eV$.

It's a good idea also to have a read up about that work energy theorem we used, it pops up quite a lot and will be quite helpful!

Will do. Thanks for the advice! :D