# Work function and wavelength

Discussion in 'Physics & Math' started by kingwinner, Mar 30, 2007.

1. ### kingwinnerRegistered Senior Member

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796
1) The maximum kinetic energy of photoelectrons is 2.8eV. When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.1 eV. What are the "work function" of the cathode and the "initial wavelength"?
K.E.initial=K.E.max=2.8eV=4.48x10^-19J
lambda.final=1.5*lambda.initial
Final energy = 1.1 eV (<---is this purely kinetic energy? why?)

K.E.max = hc/(lambda.initial) - work function
But now I have 2 unknowns: work function and lambda.initial, what can I do to solve for both?

I am stuck here...

Does anyone have any idea or insight? Any help is greatly appreciated.

Messages:
796

5. ### kevinalmRegistered Senior Member

Messages:
993
But you can state 2 equations:

2.8ev = hc/lambdai -w
1.1ev = hc/(1.5*lambdai) -w

w is the same in both equations, the ionization energy for the cathode. Between the two equations solve for either lambdai or w, then plug that result into either equation to solve for the remaining. Simple algebra.

>>edit Of course you need to use consistant units of measurement.