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(B) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function and $K_{\max} = eV_0$ ($V

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$4 \lambda$

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(B) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function and $K_{\max} = eV_0$ ($V_0$ is the stopping potential).For the first case: $\frac{hc}{\lambda} = \phi + ex$  --- $(1)$For the second case: $\frac{hc}{2\lambda} = \phi + e(\frac{x}{3})$  --- $(2)$Multiply equation $(2)$ by $3$: $\frac{3hc}{2\lambda} = 3\phi + ex$  --- $(3)$Subtract equation $(1)$ from equation $(3)$:$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = (3\phi + ex) - (\phi + ex)$$\frac{hc}{2\lambda} = 2\phi$$\phi = \frac{hc}{4\lambda}$Since the threshold wavelength $\lambda_0$ is defined as $\phi = \frac{hc}{\lambda_0}$,we have:$\frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$$\lambda_0 = 4\lambda$.

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........

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