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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For

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$\frac{hc}{6\lambda}$

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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$,where $K_{max} = eV_s$ ($V_s$ is the stopping potential).For the first case: $\frac{hc}{\lambda} = 5eV_0 + \phi$  --- $(1)$For the second case: $\frac{hc}{3\lambda} = eV_0 + \phi$  --- $(2)$Multiply equation $(2)$ by $5$: $\frac{5hc}{3\lambda} = 5eV_0 + 5\phi$  --- $(3)$Subtract equation $(1)$ from equation $(3)$:$\frac{5hc}{3\lambda} - \frac{hc}{\lambda} = (5eV_0 + 5\phi) - (5eV_0 + \phi)$$\frac{5hc - 3hc}{3\lambda} = 4\phi$$\frac{2hc}{3\lambda} = 4\phi$$\phi = \frac{2hc}{12\lambda} = \frac{hc}{6\lambda}$

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

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