NEET 2020 Physics Question 157 Solution

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is: $$K_{\text{max}} = h\nu - \phi_0 = h(\nu - \nu_0)$$ For any photoelectrons to be emitted, the energy of the incident photon must be at least equal to the work function ($$h\nu \ge h\nu_0 \implies \nu \ge \nu_0$$).

The initial frequency of incident light is: $$\nu = 1.5 \nu_0$$ When the frequency is halved: $$\nu' = \frac{\nu}{2} = \frac{1.5\nu_0}{2} = 0.75\nu_0$$

Notice that the new frequency $$\nu' = 0.75\nu_0$$ is strictly LESS than the threshold frequency $$\nu_0$$: $$\nu' < \nu_0$$ When the frequency is below the threshold frequency, the photon energy is insufficient to liberate even a single electron from the surface.

Intensity corresponds to the number of incident photons per second. Since each individual photon carries energy less than the work function ($$h\nu' < \phi_0$$), increasing the number of photons cannot cause emission. Hence, no photoelectrons are ejected, and the photoelectric current drops to exactly ZERO.