GR9677 #42


Problem


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Quantum Mechanics}Photoelectric Effect
Recall the photoelectric equation relating the incident electromagnetic wave to the kinetic energy and the work function , as in choice (B).


Alternate Solutions 
cczako 20131018 15:36:40  Remember KE=Ephi where E is the energy of the light and phi is the workfunction. Don't forget that E=hc/lambda and that hc can be written as 1240 nm*eV. That means that 500 nm is about 2.5 eV so KE is simply 2.5 2.28 which is 0.22 eV or choice B.  

Comments 
cczako 20131018 15:36:40  Remember KE=Ephi where E is the energy of the light and phi is the workfunction. Don't forget that E=hc/lambda and that hc can be written as 1240 nm*eV. That means that 500 nm is about 2.5 eV so KE is simply 2.5 2.28 which is 0.22 eV or choice B.   Dodobird 20101112 14:46:11  We get E = .12 eV, which doesn't really help us differentiate between choice A and B. There is a difference of .01 eV in 'closeness' to these answers. In fact the wording of the problem would suggest our maximum energy can only be .03 eV out of the choices given.
Help?
pam d 20110915 21:55:01 
I know this is way too late but you actually get (a reasonable mental calculation). The difference between this and the work function gives the maximum possible KE, which puts you at about .

  CarlBrannen 20101007 14:14:11  Use hbar c = 1240 nm eV. Converting 500nm, we get 2.48 eV. Subtracting 2.28 eV for the work function, we find 0.2 eV, answer (B).   vortex 20051205 15:50:02  As a twist of fate!, the formula is given in question 27.  

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