Q.Using the idea of the work function of a metal, explain why photoelectric emission from a given surface occurs only when the frequency of the incident light is above a certain minimum (threshold) value, no matter how intense the light is.

Work function as an energy threshold. The work function $\phi_0$ is the minimum energy an electron needs to be supplied, in one go, to escape the metal. In the photon picture, an electron receives energy only from a SINGLE photon at a time -- it cannot pool the energy of several photons together (at ordinary light intensities), because each photon-electron interaction is a separate, essentially instantaneous event.

Why intensity cannot compensate for low frequency. A photon of frequency $\nu$ carries a FIXED energy $h\nu$, set entirely by its frequency, regardless of the light's intensity. If $h\nu<\phi_0$, then even though increasing the intensity increases the NUMBER of such (too-weak) photons arriving per second, every individual photon is still too weak, on its own, to free any electron -- so the photocurrent stays exactly zero, however intense the light. Only once $\nu\ge\nu_0=\phi_0/h$ does each individual photon carry enough energy, and emission becomes possible; increasing the intensity beyond that point then simply increases the number of successful emissions per second (the photocurrent), as Law 1 states.

Start from the one-photon-one-electron rule, note that photon energy depends only on frequency, and conclude that intensity (photon number) cannot substitute for insufficient per-photon energy.

• Arguing that a high-intensity beam below threshold should EVENTUALLY free an electron by 'piling up' energy -- this is exactly the (wrong) classical intuition; in the photon picture, sub-threshold photons never combine their energy.
• Forgetting to state explicitly that $\nu_0=\phi_0/h$ follows directly from setting $K_{max}=0$ in Einstein's equation.