To solve this problem, we will use the photoelectric effect equation.$\\$The equation for the kinetic energy of the ejected electron is given by:$\\ K.E. = h \nu - \phi \\$where $\phi$ is the work function of the material, and $\nu$ is the frequency of the incident light.$\\$The work function $\phi$ can be calculated using the threshold wavelength $\lambda_0: \\ \phi = \frac{h c}{\lambda_0} \\$Given:$\\ \lambda_0 = 3250 \times 10^{-10}$ m$\\ \lambda = 2536 \times 10^{-10}$ m$\\ h = 4.14 \times 10^{-15}$ eVs$\\ c = 3 \times 10^8$ m/s$\\$First, calculate the work function $\phi: \\ \phi = \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3250 \times 10^{-10}} \\ \phi = \frac{12.42 \times 10^{-7}}{3250 \times 10^{-10}} \\ \phi = \frac{12.42}{3250} \times 10^3 \\ \phi = 3.82$ eV$\\$Next, calculate the energy of the incident light $E: \\ E = \frac{h c}{\lambda} \\ E = \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{2536 \times 10^{-10}} \\ E = \frac{12.42 \times 10^{-7}}{2536 \times 10^{-10}} \\ E = \frac{12.42}{2536} \times 10^3 \\ E = 4.90$ eV$\\$Now, calculate the kinetic energy $K.E.$ of the ejected electron:$\\ K.E. = E - \phi \\ K.E. = 4.90 - 3.82 \\ K.E. = 1.08$ eV$\\$Convert the kinetic energy from eV to joules:$\\ 1$ eV$= 1.6 \times 10^{-19}$ J$\\ K.E. = 1.08 \times 1.6 \times 10^{-19} \\ K.E. = 1.728 \times 10^{-19}$ J$\\$The kinetic energy is also given by $\frac{1}{2} m v^2,$ where $m$ is the electron mass $(9.11 \times 10^{-31}$ kg$): \\ \frac{1}{2} m v^2 = 1.728 \times 10^{-19} \\ v^2 = \frac{2 \times 1.728 \times 10^{-19}}{9.11 \times 10^{-31}} \\ v^2 = \frac{3.456 \times 10^{-19}}{9.11 \times 10^{-31}} \\ v^2 = 3.794 \times 10^{11} \\ v = \sqrt{3.794 \times 10^{11}} \\ v \approx 6.16 \times 10^5$ m/s$\\$Therefore, the velocity of the ejected electron is approximately $6 \times 10^5$ m/s, which corresponds to Option 1.