Threshold energy=e*c/lamda
dimensions are a red herring here.
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?
1 answer
1 answer