Final answer:
The cut-off wavelength for zinc is approximately 286 nm. The lowest frequency of light incident on zinc that releases photoelectrons from its surface is approximately 1.04 × 10^15 Hz. The maximum kinetic energy of ejected photoelectrons when photons of energy 5.50 eV are incident on zinc is approximately 1.90 × 10^-19 J.
Step-by-step explanation:
(a) Cut-off wavelength:
To determine the cutoff wavelength for zinc, we need to use the relation between energy and wavelength for a photon:
E = hc/λ
Where E is the energy, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength.
Given that the work function for zinc is 4.31 eV, we can convert it to joules:
Work function (W) = 4.31 eV × 1.60 × 10^-19 J/eV = 6.90 × 10^-19 J
Setting the energy of the photon equal to the work function, we can solve for the cut-off wavelength:
E = hc/λ = W
λ = hc/W
Plugging in the values, we get:
λ = (6.626 × 10^-34 J·s × 3.00 × 10^8 m/s) / (6.90 × 10^-19 J) ≈ 286 nm
(b) Lowest frequency:
The lowest frequency of light incident on zinc that releases photoelectrons from its surface can be determined using the relation between energy and frequency:
E = hf
Given that the work function for zinc is 4.31 eV, we can convert it to joules:
Work function (W) = 4.31 eV × 1.60 × 10^-19 J/eV = 6.90 × 10^-19 J
Setting the energy of the photon equal to the work function, we can solve for the frequency:
E = hf = W
f = E/h = W/h
Plugging in the values, we get:
f = (6.90 × 10^-19 J) / (6.626 × 10^-34 J·s) ≈ 1.04 × 10^15 Hz
(c) Maximum kinetic energy:
The maximum kinetic energy of ejected photoelectrons can be determined using the relation:
KEmax = Ephoton - Work function
Given that photons of energy 5.50 eV are incident on zinc, we can convert it to joules:
Energy of photons = 5.50 eV × 1.60 × 10^-19 J/eV = 8.80 × 10^-19 J
Plugging in the values, we get:
KEmax = (8.80 × 10^-19 J) - (6.90 × 10^-19 J) = 1.90 × 10^-19 J