# The Photoelectric Effect Help

1. May 17, 2008

### sodr2

1. The problem statement, all variables and given/known data
For an incoming light shining on a metal surface, I am given the wavelength, the cutoff voltage, the frequency (by using f = c/w). From here, Im having trouble finding the maximun Ek of the ejected electrons in electron volts (eV) and joules (J).

2. Relevant equations
Ek max = eV

E = hƒ = hc / λ

Ek max = E (photon) - work

3. The attempt at a solution

Im guessing that i can use the equation Ek max = eV by substituting in the cutoff voltage and the charge on an electron to get the Ek max in joules, but then how do I get it in electron volts? Is there a conversion factor, or another equation I use??
PLz help..

2. May 17, 2008

### rosstheboss23

Joules to eV...1.602 times 10^-19J= 1 ev

3. May 17, 2008

### sodr2

ok, so if i use Ek = eV to find the energy in joules and divide by e (1.602x10^-19), then i just end up at volts again. If the cutoff voltage is equal to the energy in eV then Id be fine, but should i be using Ek = eV ?

4. May 18, 2008

### alphysicist

If you calculate the energy term eV you can get either electron volts or Joules depending on what units you use for the charge on the electron.

If you use charge in Coulombs (for example $e= 1.6\times 10^{-19}$ C) then the energy is in Joules.

If your charge is just the number of elementary charges (here 1), then multiplying by the cutoff voltage you get electronvolts.

Two examples: If an electron moves through a 20 V potential difference, the change in potential energy is 20 electronvolts (could be positive or negative depending on the direction). If a helium nuclei (with a positive charge equal to 2e) moves through a 30 V potential difference, the change in potential energy is 60 electronvolts (again, could be positive or negative).

(By the way, I don't understand what you mean in your last post when you say you found the energy in Joules, divided by the conversion factor and got volts again. If your kinetic energy in Joules was (for example) $2\times 10^{-17} J$, then converting to electron volts would be:

[tex]
2\times 10^{-17} \mbox{ J} \frac{1 \mbox{ eV}}{1.6\times 10^{-19} \mbox{ J}} = 125 \mbox{ eV}
[/itex]

5. May 18, 2008

### sodr2

i meant that if i have a cutoff voltage for a light (lets say its 2 V) i should multiply it by e to get the energy in joules (which would be in this case 3.2 x 10^-19 J). if i take that energy and divide it by e, then ill just end up at volts again...

cause E = eV

6. May 18, 2008

### alphysicist

But you're not dividing by the charge e, you're dividing by a conversion factor. You're not ending up with volts again, you're ending up with electronvolts; but since the numerical value of this conversion factor happens to be the same as the charge on one electron, the potential energy (in eV) and the potential difference (in V) have the same numerical value in this case. That "coincidence" is why electronvolts is such a useful unit.

(My example was if an electron goes through a potential difference of V=20 volts, its potential energy change is $\Delta U=$ 20 electronvolts.)

7. May 18, 2008

### sodr2

ohh ok, thx for your help.