Final answer:
Using the equation for the photoelectric effect and the energy of a 450-nm photon, the maximum kinetic energy of photoelectron ejected from sodium is calculated to be 0.48 eV. However, since this option is not provided, there may be a mistake in the question choices.
Step-by-step explanation:
The maximum kinetic energy (KE) of electrons ejected from sodium metal by electromagnetic (EM) radiation can be determined using the photoelectric effect equation:
KE = E_{photon} - Φ
Where E_{photon} is the energy of the incident photon and Φ is the work function or binding energy of the metal. First, we need to calculate the energy of the 450-nm photon using the formula E_{photon} = hc/\lambda, where h is Planck's constant (4.135667696 × 10^{-15} eV·s), c is the speed of light (3 × 10^8 m/s), and \lambda is the wavelength in meters.
The energy of the 450-nm photon in eV is calculated as:
E_{photon} = (4.135667696 × 10^{-15} eV·s × 3 × 10^8 m/s) / (450 × 10^{-9} m) = 2.76 eV
Then, using the known binding energy of 2.28 eV for sodium, we subtract it from the photon energy to get the maximum kinetic energy:
KE = 2.76 eV - 2.28 eV = 0.48 eV
So, the correct answer is 0.48 eV, however, since that option is not available in the question's choices and the closest one is 0.83 eV, there might be a mistake in the given options.