- #1
nichelfish
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Hey guys, thanks for any help in advance. I'm working through some exercises in University Physics 13th Edition and have an issue with a photoelectric effect question.
I'm given a graph that represents stopping potential as a function of frequency of light falling onto the surface (attached is a drawing, didn't want to post a direct pic incase it went against the rules). At a stopping potential of 0, the frequency is 1.25x10^15 Hz. As in this case hf=work function (and using h=6.626x10^-34 and e as 1.602x10^-19) I get, using parenthesis to try and make it clearer:
(1.25x10^15)x(6.626x10^-34) = 8.2825x10^-19 V
Converting to eV
(8.2825x10^-19) / (1.602x10^-19) = 5.17 eV.
The answer listed in the textbook gives 4.8eV. I've tried calculations for other points along the line (larger frequencies and stopping potentials) and get the same answer.
Anybody able to hint at as to where I may be going wrong?
Thanks in advance for any help.
I'm given a graph that represents stopping potential as a function of frequency of light falling onto the surface (attached is a drawing, didn't want to post a direct pic incase it went against the rules). At a stopping potential of 0, the frequency is 1.25x10^15 Hz. As in this case hf=work function (and using h=6.626x10^-34 and e as 1.602x10^-19) I get, using parenthesis to try and make it clearer:
(1.25x10^15)x(6.626x10^-34) = 8.2825x10^-19 V
Converting to eV
(8.2825x10^-19) / (1.602x10^-19) = 5.17 eV.
The answer listed in the textbook gives 4.8eV. I've tried calculations for other points along the line (larger frequencies and stopping potentials) and get the same answer.
Anybody able to hint at as to where I may be going wrong?
Thanks in advance for any help.
