Asked by Kid
For a rectangular metal surface with dimensions 5 cm by 3 cm, the threshold wavelength for the photoelectric emission of electrons is 246.0 nm.
(a) Calculate the work function of the metal surface.
So I planned to use: hf - workfunction = eV
Then I replace f with c/lambda, since c = (f)(lambda)
So I have:
(4.135*10^-15 eV*s)(3.00*10^8 m/s)/246.0*10^-9 m - workfunction = eV
Clearly I am missing something or a lot of things, so how am I supposed to solve for the workfunction using the given information about the dimensions of the metal surface? Are the dimensions even important and if so, how?
(a) Calculate the work function of the metal surface.
So I planned to use: hf - workfunction = eV
Then I replace f with c/lambda, since c = (f)(lambda)
So I have:
(4.135*10^-15 eV*s)(3.00*10^8 m/s)/246.0*10^-9 m - workfunction = eV
Clearly I am missing something or a lot of things, so how am I supposed to solve for the workfunction using the given information about the dimensions of the metal surface? Are the dimensions even important and if so, how?
Answers
Answered by
bobpursley
Threshold energy=e*c/lamda
dimensions are a red herring here.
dimensions are a red herring here.
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