Final answer:
For light with a wavelength of 303 nm, which has an energy of approximately 4.10 eV, only lithium and beryllium will exhibit the photoelectric effect, as their work functions are less than 4.10 eV. Mercury will not, as its work function is higher at 4.50 eV.
Step-by-step explanation:
The student has asked which of the metals lithium, beryllium, and mercury will exhibit a photoelectric effect when exposed to light with a wavelength of 303 nm. To determine this, we calculate the energy of the incoming photons using the equation E = hc/\(λ\), where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3.00 x 10^8 m/s), and \(λ\) is the wavelength of the light (303 nm or 303 x 10^-9 m).
Calculating the energy gives us E = (6.626 x 10^-34 Js)(3.00 x 10^8 m/s) / (303 x 10^-9 m), which simplifies to approximately 4.10 eV. Any metal with a work function less than this energy will exhibit the photoelectric effect. In this case, all three metals, lithium (2.30 eV), beryllium (3.90 eV), and mercury (4.50 eV), are under consideration. Only lithium and beryllium, with their work functions less than 4.10 eV, will exhibit the photoelectric effect with 303 nm light. Mercury, with a higher work function of 4.50 eV, will not.