Question Detail

According to the photoelectric effect, the maximum kinetic energy (KE_max) of ejected electrons is given by:KE_max = hf - ΦWhere:h = Planck's constant = 6.625 × 10⁻³⁴ J.sf = incident frequency = 3.0 × 10¹⁵ HzΦ = work function = 2.3 eVFirst, calculate the energy of the incident photon (hf):hf = (6.625 × 10⁻³⁴ J.s) × (3.0 × 10¹⁵ Hz) = 19.875 × 10⁻¹⁹ JNext, convert the work function from eV to Joules:Φ = 2.3 eV × (1.602 × 10⁻¹⁹ J/eV) = 3.6846 × 10⁻¹⁹ JNow, calculate KE_max in Joules:KE_max = 19.875 × 10⁻¹⁹ J - 3.6846 × 10⁻¹⁹ J = 16.1904 × 10⁻¹⁹ JFinally, convert KE_max back to electron-volts (eV):KE_max = (16.1904 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV) ≈ 10.106 eVThe closest option is 10.13 eV.