To solve this problem, we will use the photoelectric effect equation.$\\$The photoelectric equation is given by:$\\ E = h \nu = \frac{hc}{\lambda} = \phi + eV \\$Where:$\\$• $E$ is the energy of the incident photon.$\\$• $h$ is Planck's constant.$\\$• $\nu$ is the frequency of the incident light.$\\$• $\lambda$ is the wavelength of the incident light.$\\$• $\phi$ is the work function of the metal.$\\$• $e$ is the charge of an electron.$\\$• $V$ is the stopping potential.$\\$Given:$\\$1. For wavelength $\lambda,$ the stopping potential is $V. \\$2. For wavelength $2\lambda,$ the stopping potential is $\frac{V}{4}. \\$Using the photoelectric equation for the first case:$\\ \frac{hc}{\lambda} = \phi + eV \quad$(Equation 1)$\\$For the second case:$\\ \frac{hc}{2\lambda} = \phi + \frac{eV}{4} \quad$(Equation 2)$\\$Subtract Equation 2 from Equation 1:$\\ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = eV - \frac{eV}{4} \\$Simplify the left side:$\\ \frac{hc}{\lambda} \left(1 - \frac{1}{2}\right) = \frac{eV(4 - 1)}{4} \\ \frac{hc}{2\lambda} = \frac{3eV}{4} \\$Rearrange to find $\frac{hc}{\lambda}: \\ \frac{hc}{\lambda} = \frac{3eV}{2} \\$Substitute back into Equation 1:$\\ \frac{3eV}{2} = \phi + eV \\$Rearrange to find $\phi: \\ \phi = \frac{3eV}{2} - eV \\ \phi = \frac{eV}{2} \\$The threshold wavelength $\lambda_0$ is given by:$\\ \phi = \frac{hc}{\lambda_0} \\$Substitute $\phi = \frac{eV}{2}: \\ \frac{eV}{2} = \frac{hc}{\lambda_0} \\$Rearrange to find $\lambda_0: \\ \lambda_0 = \frac{2hc}{eV} \\$We know from Equation 1:$\\ \frac{hc}{\lambda} = \phi + eV = \frac{eV}{2} + eV = \frac{3eV}{2} \\$Thus, $\frac{hc}{\lambda} = \frac{3eV}{2} \\$Substitute into the expression for $\lambda_0: \\ \lambda_0 = \frac{2}{3} \lambda \cdot \lambda \\ \lambda_0 = 3\lambda \\$Therefore, the threshold wavelength for the metallic surface is $3\lambda. \\$This corresponds to Option 2.