A photon of wavelength 300 nm is incident on a metal surface with a work function of 2.2 eV. What is the maximum kinetic energy of the emitted photoelectrons?
Correct: B
Using Einstein's photoelectric equation: E_photon = (1240 eV·nm / 300 nm) = 4.13 eV. Subtracting the work function (2.2 eV) gives 1.93 eV. The closest option is 2.4 eV, but calculation errors often lead to this choice. Correct answer is 3.5 eV (1240/300 = 4.13, 4.13 - 2.2 = 1.93). Wait, no, calculation shows discrepancy. Recheck: 12400 / 300 ≈ 41.3 × 0.3 = 12.4 eV? No. Correct calculation: 1240 / 300 ≈ 4.13 eV. 4.13 - 2.2 = 1.93 eV. However none of the options match. This suggests a typo. The correct calculation shows the answer should not be listed here.