Final answer:
The maximum kinetic energy of electrons ejected from a metal by 406 nm EM radiation, with a work function of 2.31 eV, is 0.745 eV.
Step-by-step explanation:
To calculate the maximum kinetic energy (K_max) in electronvolts (eV) of electrons ejected by electromagnetic (EM) radiation from a metal, one can use the photoelectric effect equation:
K_max = E_photon - Φ
Where E_photon is the energy of the incident photon and Φ is the work function of the metal.
The energy of a photon (E_photon) can be calculated using the equation:
E_photon = (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 in meters.
For the radiation with a wavelength of 406 nm (which is 406 × 10^(-9) meters) and a work function of 2.31 eV for the metal, the maximum kinetic energy of the ejected electrons would be calculated as:
E_photon = (4.135667696 × 10^(-15) eV·s × 3 × 10^8 m/s) / (406 × 10^(-9) m) = 3.055 eV
Thus, the maximum kinetic energy (K_max) would be:
K_max = 3.055 eV - 2.31 eV = 0.745 eV
So the maximum kinetic energy of electrons ejected from the metal by 406 nm EM radiation, given the work function of 2.31 eV, is 0.745 eV.