A) photoelectrons emitted from C
B) photoelectrons are emitted from A
C) photoelectrons emitted from B
D) photoelectrons emitted from all the surfaces
Correct Answer: D
Solution :
Work function (W) of a metal is defined as the minimum energy required for the emission of photoelectrons from a metal. From Einsteins equation \[E=\frac{hc}{\lambda }\]where h is Plancks constant, c is speed of light, \[\lambda \] the wavelength. Given, \[h=6.6\times {{10}^{-34}}Js,\,\,c=3\times {{10}^{8}}m/s\] \[\lambda =4000\,\overset{o}{\mathop{A}}\,=400\times {{10}^{-10}}m\] \[\therefore \] \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4000\times {{10}^{-10}}}\] \[\Rightarrow \] E = 3.09 eV Since \[E>{{W}_{A}},{{W}_{B}},{{W}_{C}}\] hence photoelectrons will be emitted from all the three surfaces.
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