Question Detail
Question
Find the maximum kinetic energy of electrons ejected from a certain material, if the material's work function is 2.3 eV and the frequency of the incident radiation is 3.0 × 10^15 Hz. (Take Planck's constant h = 6.625 × 10^-34 Js; electronic charge e = 1.602 × 10^-19 C; 1 eV = 1.602 × 10^-19 J).
Correct Answer
Option A is the correct answer.
Detailed Explanation
According to Einstein's photoelectric equation, the maximum kinetic energy (KE_max) of ejected electrons is given by:KE_max = hf - φWhere:h = Planck's constant = 6.625 × 10^-34 Jsf = frequency of incident radiation = 3.0 × 10^15 Hzφ = work function of the material = 2.3 eVFirst, calculate the energy of the incident photon (hf):hf = (6.625 × 10^-34 Js) × (3.0 × 10^15 Hz)hf = 19.875 × 10^-19 JConvert this energy to electronvolts (eV) using 1 eV = 1.602 × 10^-19 J:hf_eV = (19.875 × 10^-19 J) / (1.602 × 10^-19 J/eV)hf_eV ≈ 12.406 eVNow, calculate the maximum kinetic energy:KE_max = hf_eV - φ = 12.406 eV - 2.3 eVKE_max ≈ 10.106 eVRounding to two decimal places, KE_max ≈ 10.13 eV.
Hint
Use the photoelectric effect equation: KE_max = hf - φ. Ensure all energy values are in consistent units (e.g., eV).