To solve this problem, we will use the photoelectric effect equation.$\\$The kinetic energy of the emitted electrons is given by:$\\ K.E. = h\nu - h\nu_0 \\$where $h$ is Planck's constant, $\nu$ is the frequency of the incident light,$\\$and $\nu_0$ is the threshold frequency.$\\$The maximum kinetic energy is also related to the maximum velocity $v$ of the electrons:$\\ K.E. = \frac{1}{2}mv^2 \\$where $m$ is the mass of the electron.$\\$For the first case, when the frequency of the incident light is $2\nu_0: \\ K.E._1 = h(2\nu_0) - h\nu_0 = h\nu_0 \\$Thus, $\frac{1}{2}mv_1^2 = h\nu_0 \\$For the second case, when the frequency of the incident light is $5\nu_0: \\ K.E._2 = h(5\nu_0) - h\nu_0 = 4h\nu_0 \\$Thus, $\frac{1}{2}mv_2^2 = 4h\nu_0 \\$Now, we find the ratio of $v_1$ to $v_2: \\ \frac{v_1^2}{v_2^2} = \frac{h\nu_0}{4h\nu_0} = \frac{1}{4} \\$Taking the square root on both sides:$\\ \frac{v_1}{v_2} = \frac{1}{2} \\$Therefore, the ratio of $v_1$ to $v_2$ is $1:2. \\$This corresponds to Option 3.