Final answer:
The wavelength of EM radiation required to eject 2.00-eV electrons from calcium metal, given a binding energy of 2.71 eV, is 263 nm, which is in the ultraviolet (UV) range.
Step-by-step explanation:
To determine the wavelength of EM radiation that can eject electrons with a kinetic energy of 2.00 eV from calcium metal with a binding energy of 2.71 eV, we must first calculate the total energy needed to overcome the binding energy. We do this by summing the kinetic energy of the ejected electron and the binding energy:
Total energy = Kinetic energy + Binding energy
Total energy = 2.00 eV + 2.71 eV
Total energy = 4.71 eV
This total energy corresponds to the photon energy of the EM radiation needed to eject the electron. Using the relation between energy (E) and wavelength (λ) of a photon, E = hc/λ, where h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light in a vacuum (2.998 x 108 m/s), and λ is the wavelength in meters, we can calculate the wavelength. First, we must convert the energy from electron volts to joules (1 eV = 1.602 x 10-19 J).
E (in joules) = 4.71 eV * 1.602 x 10-19 J/eV = 7.54 x 10-19 J
Now, we can calculate the wavelength:
λ = hc/E
λ = (6.626 x 10-34 J·s)(2.998 x 108 m/s) / (7.54 x 10-19 J)
λ = 2.63 x 10-7 m or 263 nm
This wavelength corresponds to ultraviolet (UV) radiation, falling below the visible light spectrum.