Q.(a) Why is the frequency and not the intensity of light source that determines whether emission of photoelectrons will occur or not? Explain.

Each photon interacts with one electron and must individually carry enough energy (hν) to overcome the work function — more intensity just means more photons, not more energy per photon; solving Einstein's equation gives v ≈ 1.05×10⁶ m/s.

1. Why frequency, not intensity, determines emission: In the photon picture of light, radiation of frequency $\nu$ consists of a stream of photons, each carrying a fixed quantum of energy $E = h\nu$. Photoemission is a one-photon–one-electron process: a single photon is absorbed by a single electron in the metal, and emission can occur only if that photon's energy $h\nu$ is at least equal to the work function $\phi_0$ of the metal (the minimum energy needed to free an electron from the surface). Intensity of light, on the other hand, is a measure of the number of photons striking the surface per unit time (and area) — increasing intensity increases the number of photons, and hence the number of photoelectrons emitted per second (photoelectric current), but it does not increase the energy of each individual photon. So if $h\nu < \phi_0$, no matter how intense the light (how many low-energy photons arrive), no electron can absorb enough energy in one interaction to escape — emission simply will not occur below the threshold frequency, regardless of intensity.
2. Calculating the photoelectron's velocity: Given: work function $\phi_0 = 1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$, wavelength $\lambda = 3000\ \text{Å} = 3\times10^{-7}\ \text{m}$, $h = 6.63\times10^{-34}\ \text{Js}$, $m_e = 9.1\times10^{-31}\ \text{kg}$, $c = 3\times10^8\ \text{m/s}$. Energy of incident photon: …