• # question_answer Light of wavelength 4000 Å falls on a photosensitive metal and a negative 2V potential stops the emitted electrons. The work function of the material (in eV) is approximately             $(h=6.6\times {{10}^{-34}}Js,\ \ e=1.6\times {{10}^{-19}}C,\ \ c=3\times {{10}^{8}}m{{s}^{-1}})$ A) 1.1 B) 2.0 C) 2.2 D) 3.1

 [a] Energy of incident light $E\,(eV)=\frac{12375}{4000}=3.09\,eV$ Stopping potential is  2V  so ${{K}_{\max }}=2\,eV$ Hence by using $E={{W}_{0}}+{{K}_{\max }}$; W0$=1.09\,eV$$\approx \,1.1\,eV$