question_answer

Figure shows variation of stopping potential \[({{V}_{0}})\] with the frequency \[\left( \mathbf{v} \right)\] for two photo- sensitive materials\[{{M}_{1}}\] and \[{{M}_{2}}\]. (i) Why is the slope same for both lines? (ii) For which material will the emitted electrons have greater kinetic energy for the incident radiations of the same frequency? Justify your answer.                        

Answer:

                (i) The slope \[{{V}_{0}}-v\] graph gives the value of \[\frac{h}{e},\] which is same for both materials \[{{M}_{1}}\]and \[{{M}_{2}}\]. (ii) According to Einstein's photoelectric equation, \[{{K}_{\max }}=hv-{{W}_{0}}=hv-h{{v}_{0}}\] The material A^ has a lower value of threshold frequency\[{{v}_{0}}\]. So \[{{M}_{1}}\] will emit photoelectrons of greater kinetic energy for the same frequency \[v\] of the incident radiation.