1) The graph of stopping potential Vs against frequency v of incident radiation is plotted for two different metals P and Q as shown in the graph, $\phi_{P}$ and $\phi_{Q}$ are work-functions of P and Q respectively, then A) $\phi_{P} > \phi_{Q}$ B) $\phi_{P} < \phi_{Q}$ C) $\phi_{P} = \phi_{Q}$ D) $v_{0} ' < v_{0}$ Answer: Option BExplanation: The work function of surface $\phi = h v_{0}$ where h= Planck's comnstant and v0= threshold frequency From graph it si clear that (v0)P < (v0) Q $\therefore$ $\phi_{P} < \phi_{Q}$
