Work function of a metal surface is 2eV. Maximum kinetic energy of photoelectrons emitted from the surface for incidence of light of wavelength `4140 Å` is 1eV. What is the threshold wavelength of radiation for that surface?

Explanation

From Einstein's photoelectric equation,
`E_("max") = hf - W_(0)`
or, `hf = E_("max") + W_(0) = 1 + 2 = 3eV`
Now, `(hf)/(W_(0)) = (hf)/(hf_(0)) = (f)/(f_(0)) = (c//lamda)/(c//lamda_(0)) = (lamda_(0))/(lamda)`
`:. lamda_(0) = lamda (hf)/(W_(0)) = 4140 xx (3)/(2) = 6210 Å`