How Does Photon Wavelength Affect Electron Ejection and Metal Work Function?

[science.girl]

Homework Statement

A photon of wavelength 250 nm ejects an electron from a metal. The ejected electron has a de Broglie wavelength of 0.85 nm.

(a) Calculate the kinetic energy of the electron.

(b) Assuming that the kinetic energy found in (a) is the maximum kinetic energy that it could have, calculate the work function of the metal.

(c) The incident photon was created when an atom underwent an electronic transition. On the energy level diagram of the atom (given by link), the transition labeled X corresponds to a photon wavelength of 400 nm. Indicate which transition could be the source of the original 250 nm photon by circling the correct letter.

Diagram is last page here: http://apcentral.collegeboard.com/apc/public/repository/ap09_frq_physics_b.pdf

Homework Equations

KE = .5mv^2

The Attempt at a Solution

a) The kinetic energy of an electron:
f = c/\lambda
f = (3.0 * 10^8 m/s)/(8.5 * 10^(-10) m)
f = 3.53 * 10^17 Hz

E = hf
E = (4.14 * 10^(-15) eV\bullets)(3.53 * 10^17 Hz)
E = 1461.42 eV

b) \phi = Work Function
KE_max = hf - \phi
1461.42 eV = 1461.42 eV - \phi
\phi = 0

But I know this is incorrect.

Would I have to use KE_max = e\DeltaV for part (a)? If so, how do I find \DeltaV?

 

[rl.bhat]

The kinetic energy of an electron:
f = c/LaTeX Code: \\lambda
f = (3.0 * 10^8 m/s)/(8.5 * 10^(-10) m)
f = 3.53 * 10^17 Hz
This is wrong.
From de Broglie wave length you can find momentum first. p = h/λ.

 

[diazona]

For part (a), you calculated the energy of the photon, not of the electron. Remember that the electron starts out "attached" (bound) to the atom, and it takes part of the photon's energy just to separate the electron from the atom. (That part is called the work function.) The leftover energy stays with the electron as kinetic energy, and it's that leftover energy that you are asked to find in part (a). You can do it by using the electron's de Broglie wavelength of 0.85 nm. Do you know a formula involving de Broglie wavelength?

 

[science.girl]

So, using p = h/\lambda :

p = (6.63 * 10^(-34) J*s)/(8.5*10^(-10) m)
p = 7.8 * 10^(-25)

Or would you use h = 4.14 * 10^(-15) eV*s?
In which p = 4.87*10^(-6)?

And what are the units for the momentum? Simply kg*m/s?

 

[rl.bhat]

And what are the units for the momentum? Simply kg*m/s?
Yes.
Now using momentum find energy. E = p^2/2m

 

[science.girl]

And what are the units for the momentum? Simply kg*m/s?
Yes.
Now using momentum find energy. E = p^2/2m

E = [(4.87*10^(-6) kg*m/s)^2]/[2(9.11*10^(-31) kg)] = 1.30 * 10^19 J

If this is correct, than for part (b), E = hf - \phi, correct? Or is there another equation?

 

[rl.bhat]

E = [(4.87*10^(-6) kg*m/s)^2]/[2(9.11*10^(-31) kg)] = 1.30 * 10^19 J

If this is correct, than for part (b), E = hf - \phi, correct? Or is there another equation?

p should be in kg.m/s not in eV.

 

[science.girl]

p should be in kg.m/s not in eV.

Oops! p = 7.8 * 10^(-25) kg.m/s
Therefore:
E = [(7.8 * 10^(-25) kg.m/s)^2]/[2(9.11*10^(-31) kg)] = 3.38*10^(-19) J

And E = hf - \phi?

UPDATE:
Is this it?

E = hf - \phi = hc/\lambda - \phi where you take the difference of the energy of the electron and the energy of the photon to reach the answer?

 

[science.girl]

Never mind. I figured it out. Thanks for your help.