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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the wor

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$4\lambda$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \phi_0$,where $\phi_0$ is the work function.Case $(i)$: For wavelength $\lambda$,stopping potential is $3V_0$:$3eV_0 = \frac{hc}{\lambda} - \phi_0$  ......... $(1)$Case $(ii)$: For wavelength $2\lambda$,stopping potential is $V_0$:$eV_0 = \frac{hc}{2\lambda} - \phi_0$  ......... $(2)$Multiply equation $(2)$ by $3$:$3eV_0 = \frac{3hc}{2\lambda} - 3\phi_0$  ......... $(3)$Equating $(1)$ and $(3)$:$\frac{hc}{\lambda} - \phi_0 = \frac{3hc}{2\lambda} - 3\phi_0$$2\phi_0 = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$$\phi_0 = \frac{hc}{4\lambda}$Since the threshold wavelength $\lambda_0 = \frac{hc}{\phi_0}$,substituting $\phi_0$:$\lambda_0 = \frac{hc}{hc / 4\lambda} = 4\lambda$.

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