# Work function for a certain metal is 1.5 eV, what is its stopping potential ? stuck!

1. Apr 20, 2006

### mr_coffee

Hello everyone! Our professor told us how to do this problem but it doesn't seem to be working. Here is what the problem says:
(a) If the work function for a certain metal is 1.5 eV, what is its stopping potential for electrons ejected from the metal when light of wavelength 337 nm shines on the metal?

Well he explained, the theory behind it, and it made sense. He then said, the Kenetic Engery (KE) will turn equal the Stopping Potential * electric charge of an electron.

THe KE is in electron volts, eV.
KE = (Stopping Potential)(Eelctric charge of electron);
We want the stopping potential so:
(stoppping Poential) = KE/1.602E-19;

We find KE by the following:
h = planks constant = 4.136E-15 eVs;
KE = h*f - work;
where f is the frequency;
We find f by:
f = c/wave length;
f = (3E8 m/s)/(337E-9m) = 8.902E14 s^-1

they give us the work, so all we need now is to solve for KE, then we have to divide KE by charge of an electoron and that should be the answer.

KE = (4.136E-15 eVs)( 8.902E14 s^-1) - 1.5eV;
KE = 2.1819 eV;
PE = 2.1819eV/1.602E-19;
PE = 1.36196E19;

Which is wrong. Someone got the answer right, but their answer was alot smaller tahn mine. They got 2.302 somthing, had no E19. Any ideas where i messed up? Thanks!

they want the answer in Volts.
I submitted 1.3619 omitting the E19, and also tried putting E19, both don't work.

Last edited: Apr 20, 2006
2. Apr 20, 2006

### Andrew Mason

The stopping potential is the potential (energy / unit charge) measured in volts (joules/coulomb) that must be applied to stop the electrons from being ejected from the surface when the light is shone on it.

If the energy of the incident photon is greater than the work required to remove the electron from the surface plus the applied (-) potential, electrons will leave the surface with some kinetic energy. The stopping potential is the applied potential that makes this KE = 0.

So the stopping potential is given by:

$$q_eV_s = E_{photon} - q_e\phi$$
$$V_s = h\nu/q_e - \phi$$

where $V_s$ is the stopping potential and $\phi$ is the work function (Joules/coulomb).

I think you understand this. Your problem may be just in math. To avoid confusion, work out the solution algebraically and the plug in numbers at the end.

In SI units:

$$h\nu/q_e = hc/\lambda q_e = 6.63e-34 * 3e8/337e-9*1.6e-19 = 3.69$$ Volts.

So V_s = 3.69 - 1.5 = 2.29 V.

AM

3. Apr 20, 2006

### mr_coffee

Thank you for the explanation, it makes senes to me. It seems to be algebra like you said. But i submitted your answer and it still didn't like it. 2.29V. I'm wondering why you used 6.663eE-34 J Hz, instead of 4.136x10^-15 eVs? I plugged in 4.136E-15 instead of 6.63E-34 and got my orginal answer whichi is also wrong. But i don't see why its saying yours is wrong.

4. Apr 20, 2006

### nrqed

He used that value because he is working in SI unites (kg, meter, second). I think it is just a simple mistake in the last step...3.69-1.5 gives 2.19, not 2.29 volt.

5. Apr 20, 2006

### mr_coffee

Whoops! you are correct again! your a machine hah, thanks again!

6. Apr 20, 2006

### nrqed

How did you know (that I am a machine)? I am actually 200 Pentium 4 connected in parallel.
:rofl: