question_answer

Sketch the graph, showing the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B having threshold frequencies \[{{v}_{0}}'\]and \[{{v}_{0}},\] respectively \[(v{{'}_{0}}>{{v}_{0}}).\]

(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 on the potential axis represent?

Answer:

            Since, \[K{{E}_{\max }}=hv-\phi \] (Einstein's photoelectric equation)             \[\Rightarrow \] \[e{{V}_{0}}=hv-\phi \] Dividing by e on both sides, we get             \[{{V}_{0}}=\left( \frac{h}{e} \right)v-\frac{\phi }{e}\] If we compare it with,      \[y=mx+c\] Then,    \[Y={{V}_{0}}\] and \[m=\frac{h}{e},\] \[x=v\] and \[c=\frac{-\phi }{e}\] Graph of stopping potential versus frequency of incident radiation is given below. Graph of stopping potential versus frequency of incident radiation (i) Work function, \[\phi =h{{v}_{0}}\Rightarrow \phi \propto {{v}_{0}}\] Higher the threshold frequency = Higher the work function. According to the question,                                     \[{{v}_{0}}'>{{v}_{0}}\] \[\therefore \] Metal A have higher work function \[(\phi ).\] (ii) By Einstein's photoelectric equation,             \[K{{E}_{\max }}=hv-\phi \] or         \[e{{V}_{0}}=hv-\phi ;\] \[{{V}_{0}}=\left( \frac{h}{e} \right)v-\frac{\phi }{e}\] \[\left[ comparing\,\,with\,\,y=mx+c\,\,and\,\,m=\frac{h}{e} \right]\] The slope of \[({{V}_{0}}\,versus\,\,v)\] graph \[=\frac{h}{e}\] constant (iii) Intercept of graph on the potential axis             \[=-\frac{\phi }{e}\]                   \[\left[ \because c=\frac{-\phi }{e} \right]\]