# Homework Help: Very simple question (Photoeletric effect)

1. Jan 18, 2010

### thefifthlord

Is eVo = Max Ek?

Vo being the cutoff potential at a given frequency and e being the elementary charge? And ofcourse Ek is max kinetic energy of ejected electrons. Also is the unit for this equation in electron volts?

Last edited: Jan 18, 2010
2. Jan 18, 2010

### Matterwave

No, the equation is K=hf-eV

Where K is the max kinetic energy, f is the frequency, h is Planck's constant, e the elementary charge, and V is the stopping voltage.

You can think of it like this. eV is the binding energy for that electron, while hf is the energy of the light coming to free that electron. So the max energy of the electron afterwords is freeing energy - binding energy. A portion of the light's energy is gone into freeing the electron, and the rest is gone into the kinetic energy of the electron.

3. Jan 18, 2010

### thefifthlord

What i mean is, i have several sources indicated that qVo is the work, however experimentally this doesn't match up, and i have several sources that indicate that qVo is the maximum kinetic energy.

My book says that Ekmax = qVo
Im only asking because a max kinetic energy of a fraction of an elementary charge seems low.

"If the potential difference is expressed in volts, this maximum kinetic energy of the electrons can be found from the relationship E = qVo" "At a sufficiently negative collector potential, Vo, the current will stop all together, at this point the retarding potential difference between the electrodes represents the maximum possible kinetic energy of the ejected electrons."

Last edited: Jan 18, 2010
4. Jan 18, 2010

### Matterwave

From a photoelectric effect viewpoint, f, the frequency of the light, determines the maximum kinetic energy of the stripped off electron, not the stopping voltage.

What your equation looks like, is if the electron is accelerated through a potential Vo.

If for example, you have a parallel plate capacitor, with the voltage between two plates being Vo, then an electron starting from rest and accelerated through that voltage will gain kinetic energy equal to eVo.

This has nothing to do with the photoelectric effect though...

5. Jan 18, 2010

### thefifthlord

Well the title of the lab is analyzing the photoelectric effect...

There is a collector and a peice of cesium inside a phototube, so i'm not sure what to deduce.

6. Jan 18, 2010

### Matterwave

Oh, I just read your quote again, I think I know what is going on.

I got confused in my previous post, sorry. It's been a while since I've done this.

Instead of E=hf-eV, it should be E=hf-W where W is the work function of the plate.

So, if I have a work function W=5eV then, it takes 5eV of energy to strip off an electron.

If then I put a negative voltage against the direction the electrons should be coming off, i get the modified equation: E=hf-W-eV where V is the voltage I apply. This is because the electron now must fight against the new voltage.

If I use some f such that hf>W, then I have excess kinetic energy. To find this kinetic energy, I can adjust V such that no more electrons come off (this is "stopping voltage" I had confused with the work function from before). If V is such that no more electrons come off, I know that E=hf-W-eV=0. Which means that eV=hf-W=Emax from before.

Sorry for the mix up. If I now vary f, and V I can find Emax for different values of f. From there, I can deduce W. This is the usual set up for the PE effect.

7. Jan 18, 2010

### thefifthlord

So it is indeed the Ek max? I was wondering why it's so low, when if i manually enter in values for the hf - W, i get something higher. Perhaps the cesium wasn't pure.

(And i had just drawn a circuit diagram lol)

Last edited by a moderator: May 4, 2017
8. Jan 18, 2010

### Matterwave

Yes, Emax = eV where V is the stopping voltage. Usually, you use this to find W, and see if it matches with known values of W (or guess what kind of material you've been hitting with light).

If your eV isn't matching up with your hf-W then there are some errors present. How far off are the values?

9. Jan 18, 2010

### thefifthlord

More than 100% off, but then again this book may not be using pure cesium thus the work function i used may have been wrong. Note i wasn't using the work function from the Ek value i got, but rather what is the accepted work function for pure cesium.

10. Jan 18, 2010

### Matterwave

Well, if you care to put up some numbers (assuming your data set isn't ginormous), I could look to see if there's anything obvious that pops out to me. But indeed, the stopping voltage is used to measure Ek in this experiment.

11. Jan 18, 2010

### thefifthlord

K thanks, btw, to anyone reading this The energy given if you use E = qVo is in Joules if you do the multiplication not Electron volts! I just wasted half an hour trying to understand why my values were so low, when infact i thought i had 4.0 * 10^-20 eV i had 4.0 * 10^-20 Joules which makes perfect sense!

12. Jan 18, 2010

### Matterwave

It's in Joules if you use q=1.602*10^-19 Coulombs.

If you use q=1e then it's in electron volts.

For example, if q=1e and V=1 volt, then E=qV yields E=1e*1volt = 1eV

But if you do E=qV and instead of q=1e, you use q=1.602*10^-19 Coulombs, then E=qV yields E=1.602*10^-19 Coulombs*1volt = 1.602*10^-19J = 1eV

Equivalent.