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nathan17
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Homework Statement
The work function of caesium is 1.35eV. A photoelectric cell has a caesium surface and is illuminated with monochromatic light of wavelength 400nm. What reverse potential difference must be applied to the tube to just stop a current passing through it?
Homework Equations
Kmax = Vs * e
E = hf = hc / λ
The Attempt at a Solution
λ = 400nm = 4.00x10-7
E = hf = hc / λ
= 6.63x10-34 * 3x108 / 4.00x10-7
= 4.97x10-19J
= 4.97x10-19 / 1.6x10-19
= 3.11 eV
Kmax = E - W
= 3.11 - 1.35
= 1.76
Kmax = Vs * e
Divide both sides by e
Vs = K / e
= 1.76 / 1.6x10-19
= 1.1x1019
Other Information
Now, the answer in the back of the book is :
Vs = 1.76 V
I got that in the second step, but I'm not sure if I worked it out correctly. Can someone go over this and check it out for me.
Cheers,
Nathan
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