- #1
mr_coffee
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Hello everyone, I'm alittle confused on why this isn't right.
THe problem asks:
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 problem asks:
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:
[tex]q_eV_s = E_{photon} - q_e\phi[/tex]
[tex] V_s = h\nu/q_e - \phi[/tex]
where [itex]V_s[/itex] is the stopping potential and [itex]\phi[/itex] 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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