To solve this problem, we will use the photoelectric effect equation:$\\ E = h \nu = \frac{hc}{\lambda} \\$The stopping potential $V_s$ is related to the maximum kinetic energy of the emitted electrons:$\\ eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \\$where $\lambda_0$ is the threshold wavelength.$\\$Given:$\\$• For wavelength $\lambda,$ the stopping potential is $3V_0: \\ e \cdot 3V_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad (1) \\$• For wavelength $2\lambda,$ the stopping potential is $V_0: \\ e \cdot V_0 = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad (2) \\$Subtract equation (2) from equation (1):$\\ e \cdot 3V_0 - e \cdot V_0 = \left(\frac{hc}{\lambda} - \frac{hc}{2\lambda}\right) \\ 2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \\ 2eV_0 = \frac{hc}{\lambda} \left(1 - \frac{1}{2}\right) \\ 2eV_0 = \frac{hc}{2\lambda} \\$From this, we can express $\frac{hc}{2\lambda}$ in terms of $V_0: \\ \frac{hc}{2\lambda} = 2eV_0 \\$Substitute $\frac{hc}{2\lambda} = 2eV_0$ into equation (2):$\\ eV_0 = 2eV_0 - \frac{hc}{\lambda_0} \\ \frac{hc}{\lambda_0} = 2eV_0 - eV_0 \\ \frac{hc}{\lambda_0} = eV_0 \\$Thus, the threshold wavelength $\lambda_0$ is:$\\ \lambda_0 = \frac{hc}{eV_0} \\$Since $\frac{hc}{2\lambda} = 2eV_0,$ we have:$\\ \lambda = \frac{hc}{4eV_0} \\$Therefore, $\lambda_0 = 4\lambda. \\$This corresponds to Option 1.