Final answer:
The maximum kinetic energy of photoelectrons when a 141-nm radiation illuminates a gold-plated electrode can be found by subtracting the work function for gold from the photon's energy, resulting in a maximum kinetic energy of 3.706 eV.
Step-by-step explanation:
To find the maximum kinetic energy of ejected photoelectrons when a 141-nm radiation illuminates a gold-plated electrode, we use the photoelectric effect equation. This equation states that the maximum kinetic energy (KEmax) of the ejected photoelectrons is the energy of the photons (hv) minus the work function (φ) of the material:
KEmax = hv - φ
First, convert the wavelength of the radiation to energy using Planck's constant (h) and the speed of light (c). The energy of a photon (E) can be calculated as:
E = hc / λ
Where:
- h is Planck's constant (4.135667696 x 10-15 eV·s)
- c is the speed of light (3 x 108 m/s)
- λ is the wavelength of the incident radiation (141 nm)
Converting the wavelength to meters gives us 141 x 10-9 m. Calculating the energy:
E = (4.135667696 x 10-15 eV·s) × (3 x 108 m/s) / (141 x 10-9 m)
E = 8.806 eV
Now, subtract the work function for gold (5.1 eV) from the photon energy:
KEmax = 8.806 eV - 5.1 eV
KEmax = 3.706 eV
The maximum kinetic energy of the ejected photoelectrons is 3.706 eV.