Question Detail
Question
The threshold wavelength of photoelectric emission of a metal is 4000 Å. Minimum energy required to eject photoelectron? (h = 6.626×10⁻³⁴ Js)
Correct Answer
Option B is the correct answer.
Detailed Explanation
E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸)/(4000×10⁻¹⁰) = (1.9878×10⁻²⁵)/(4×10⁻⁷) = 4.97×10⁻¹⁹ J. Convert to eV: (4.97×10⁻¹⁹)/(1.6×10⁻¹⁹) = 3.1 eV.
Hint
E = hc/λ.