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
KE = .5mv^2
The Attempt at a Solution
a) The kinetic energy of an electron:
f = c/[tex]\lambda[/tex]
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[tex]\bullet[/tex]s)(3.53 * 10^17 Hz)
E = 1461.42 eV
b) [tex]\phi[/tex] = Work Function
KEmax = hf - [tex]\phi[/tex]
1461.42 eV = 1461.42 eV - [tex]\phi[/tex]
[tex]\phi[/tex] = 0
But I know this is incorrect.
Would I have to use KEmax = e[tex]\Delta[/tex]V for part (a)? If so, how do I find [tex]\Delta[/tex]V?
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