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(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $v$ by the equation: $eV_s = hv - \phi$,wh

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$3v/2$

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(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is related to the frequency $v$ by the equation: $eV_s = hv - \phi$,where $\phi = hv_0$ is the work function and $v_0$ is the threshold frequency.For the first case: $e(V_0/2) = hv - \phi$  ..... $(1)$For the second case: $e(V_0) = h(v/2) - \phi$  ..... $(2)$Note: The stopping potential is given as negative,which indicates the magnitude of the potential required to stop the electrons. Thus,$eV_s = h(v - v_0)$.From $(1)$: $eV_0/2 = hv - hv_0 \Rightarrow eV_0 = 2hv - 2hv_0$From $(2)$: $eV_0 = hv/2 - hv_0$Equating the two expressions for $eV_0$:$2hv - 2hv_0 = hv/2 - hv_0$$2hv - hv/2 = 2hv_0 - hv_0$$3hv/2 = hv_0$$v_0 = 3v/2$.

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