# What is the maximum speed of the ejected electrons? good times!

Hello everyone, i'm alittle confused on why this isn't right.
What is the maximum speed of the ejected electrons?

They give you the following:
With the help of others i found:

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?
2.19 V

Andrew Mason said it best:
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)./QUOTE]

So i figured this formula would work,
Stopping Potential = 2.19 V
.5*mv^2 = Stopping Potential;
solve for V to find the speed.
V = sqrt(StoppingPotential/(.5*m));
V = 2.5E15 m/s

but it was wrong. ANy ideas what i'm missing? Thanks!

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dav2008
Gold Member
eV is a measure of energy, not voltage, so in your equation $$V_s = h\nu/q_e - \phi$$ you are equating potential to potential plus energy.

Also you set voltage equal to kinetic energy. Again, voltage is energy/charge.

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Your right, my units don't work out at all. I end up getting volts/kg hm...
I don't see how this is going to work out, are you saying I should convert my 2.9 V into somthing else?

dav2008
Gold Member
You have to realize what is going on.

Once the electron has been shot out, it takes a potential of 2.19 Joules/coulomb(if you did your calculations correctly) to stop them.

If it takes 2.19 Joules/Coulomb to stop the electron, what does that tell you about its initial kinetic energy?

dav2008
Gold Member
I guess the key thing to realize is that all this is is conservation of energy. You don't need to memorize any formulas if you just understand the situation and write a conservation of energy statement for it.

In the first case you have the energy of the photon converted into the work function energy plus the energy of the electron. Then you have the energy of the electron converted into electric potential (qV)

Oooo! i got you so it should be:
.5*mv^2 = (charge of an electron)(2.19 J/C)?