- #1
Gh0stS3C
- 8
- 2
- Homework Statement
- The ejection of the photo electron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
- Relevant Equations
- Equation used for calculating Work Function :
Energy of photon = Work function + Kinetic Energy {EQ 1}
Work Function = Energy of photon - Kinetic Energy {From EQ 1}
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Energy of photon = hc/lambda [h = Planck's Constant, c = speed of light, lambda = wavelength]
So I know exactly how to solve this question with the relevant equations. However problem I ran into was how I substitute for K.E (Kinetic Energy). I asked one of my teachers for help. She said that since the voltage applied stops the experiment, it is equivalent to the energy of the electron which she then proceeded to take as 0.35 eV (This is what I made out from her explanation, studying through online classes is proving to be very difficult for me.) Can somebody walk me through the process of how we arrived at 0.35 eV as the K.E.
My attempt so far :
Wavelength = 256.7 nm
converting to metres --> 256.7 x 10^-9 m
Energy of photon = 6.626 x 10^-34 x 3 x 10^8/256.7x10^-9
I cannot proceed further due to my doubt regarding eV conversion but I know what to do after I get the K.E value in eV.P.S Please let me know of any tips to make my question more clear. I'm new here but looking forward to learn :)
My attempt so far :
Wavelength = 256.7 nm
converting to metres --> 256.7 x 10^-9 m
Energy of photon = 6.626 x 10^-34 x 3 x 10^8/256.7x10^-9
I cannot proceed further due to my doubt regarding eV conversion but I know what to do after I get the K.E value in eV.P.S Please let me know of any tips to make my question more clear. I'm new here but looking forward to learn :)