question_answer

Sketch the graphs, showing the variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies \[v_{0}^{'}>{{v}_{0}},\] respectively. (i) Which of the two metals, A or B has higher work function? (ii) What information do you get from the slope of the graphs? (iii) What does the value of the intercept of graph 'A' on the potential axis represent?                

Answer:

                The graph is shown in Fig.                 (i) Metal a has a higher work function \[(W_{0}^{'}=hv_{0}^{'})\]because \[v_{0}^{,}>{{v}_{0}}.\]                 (ii) As \[e{{V}_{0}}=hv-{{W}_{0}}\] or \[{{V}_{0}}=\left( \frac{h}{e} \right)v-\frac{{{W}_{0}}}{e}\]                 \[\therefore \] Slope of \[{{V}_{0}}-v\]graph \[=\frac{h}{e}\]                 (iii) Intercept of graph A on the potential ? axis \[=-\frac{W_{0}^{'}}{e}=-\frac{hv_{0}^{'}}{e}\]