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(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the wor

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$V_2 > 2V_1$

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(D) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function of the metal.For wavelength $\lambda$,the stopping potential is $eV_1 = \frac{hc}{\lambda} - \phi$  --- $(1)$.For wavelength $\frac{\lambda}{2}$,the stopping potential is $eV_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ --- $(2)$.From equation $(1)$,we have $\frac{hc}{\lambda} = eV_1 + \phi$.Substituting this into equation $(2)$:$eV_2 = 2(eV_1 + \phi) - \phi$$eV_2 = 2eV_1 + 2\phi - \phi$$eV_2 = 2eV_1 + \phi$Since the work function $\phi > 0$,it follows that $eV_2 > 2eV_1$,which implies $V_2 > 2V_1$.

In an experiment of photoelectric effect,the stopping potential was measured to be $V_{1}$ and $V_{2}$ with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The relation between $V_{1}$ and $V_{2}$ is:

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