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 × 1015 Hz. (Take Planck's constant h = 6.625 × 10-34 Js; ... 1 eV = 1.602 × 10-19 J).
Correct Answer
Option A is the correct answer.
Detailed Explanation
The maximum kinetic energy (KEmax) of ejected electrons in the photoelectric effect is given by Einstein's photoelectric equation:KEmax = hf - ΦWhere:h = Planck's constant = 6.625 × 10-34 Jsf = frequency of incident radiation = 3.0 × 1015 HzΦ = work function = 2.3 eVFirst, calculate the energy of the incident photon (E = hf) in Joules:E = (6.625 × 10-34 Js) × (3.0 × 1015 Hz) = 19.875 × 10-19 JNext, convert the photon energy from Joules to electron-volts (eV), using 1 eV = 1.602 × 10-19 J:E (eV) = (19.875 × 10-19 J) / (1.602 × 10-19 J/eV) ≈ 12.406 eVNow, calculate the maximum kinetic energy:KEmax = E (eV) - Φ (eV)KEmax = 12.406 eV - 2.3 eV ≈ 10.106 eVRounding to two decimal places, KEmax ≈ 10.13 eV.
Hint
Use Einstein's photoelectric equation: KEmax = hf - Φ. Remember to ensure all energy terms are in the same units (either Joules or electron-volts) before performing calculations.