To solve this problem, we need to understand the photoelectric effect and how changes in frequency and intensity affect the photoelectric current.$\\$The photoelectric effect is described by the equation:$\\ E = h \nu = h \nu_0 + K_{max} \\$where:$\\$• $E$ is the energy of the incident photon.$\\$• $h$ is Planck's constant.$\\$• $\nu$ is the frequency of the incident light.$\\$• $\nu_0$ is the threshold frequency.$\\$• $K_{max}$ is the maximum kinetic energy of the emitted electrons.$\\$Initially, the frequency of the light is $1.5 \nu_0. \\$The energy of the incident photons is $E = h \times 1.5 \nu_0. \\$The kinetic energy of the emitted electrons is:$\\ K_{max} = h \times 1.5 \nu_0 - h \nu_0 = 0.5 h \nu_0. \\$Now, if the frequency is halved, the new frequency is:$\\ \nu' = \frac{1.5 \nu_0}{2} = 0.75 \nu_0. \\$The energy of the photons with the new frequency is:$\\ E' = h \times 0.75 \nu_0. \\$Since $\nu' < \nu_0,$ the energy of the photons is less than the work function of the material.$\\$Therefore, no electrons will be emitted, and the photoelectric current will be zero.$\\$Doubling the intensity does not affect the emission of electrons if the frequency is below the threshold.$\\$Intensity affects the number of emitted electrons, not their emission if the frequency is insufficient.$\\$Thus, the photoelectric current will be zero.$\\$This corresponds to Option 3: Zero.