Final answer:
The maximum kinetic energy of ejected photoelectrons when UV radiation falls on gold is calculated using the energy of photons and the work function of gold. By subtracting the work function from the photon's energy, we arrive at the kinetic energy, which corresponds to one of the provided options.
Step-by-step explanation:
The student's question about the maximum kinetic energy of ejected photoelectrons when UV radiation with a wavelength of 120 nm falls on gold deals with the concept of photoelectric effect, which is a fundamental principle in Physics, specifically in the field of quantum mechanics.
To calculate the maximum kinetic energy of the photoelectrons, we can use the energy of the incoming photons (E = hc/λ) and the work function of the gold metal (φ), which is the energy required to remove an electron from the metal's surface. The work function is given as 4.82 eV. The energy of a photon can be calculated using the formula E = hc/λ, where h is Planck's constant (4.135667696 × 10^-15 eV·s), c is the speed of light (approximately 3 × 10^8 m/s), and λ is the wavelength of the incident light (120 nm).
The maximum kinetic energy (KEmax) of the ejected photoelectrons is then given by the equation KEmax = E - φ. After calculating the energy E and subtracting the work function φ, we can determine KEmax. The calculated value would then match one of the provided multiple-choice options.