In a photoelectric effect experiment, light of wavelength 200 nm shines on a metal surface with work function 4.0 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Use h = 4.14 × 10^-15 eV·s and c = 3.0 × 10^8 m/s.)
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Correct answer
A. 2.2 eV
Principle or equation
Einstein's photoelectric equation: K_max = hf - φ, where f = c/λ.
Why this answer is correct
First find photon energy: E = hc/λ = (4.14 × 10^-15 eV·s × 3.0 × 10^8 m/s) / (200 × 10^-9 m) = (1.242 × 10^-6 eV·m) / (2.0 × 10^-7 m) = 6.21 eV. Then K_max = 6.21 eV - 4.0 eV = 2.21 eV, approximately 2.2 eV.
Example
If λ = 300 nm and φ = 2.0 eV, then E = 4.14 eV, K_max = 2.14 eV.
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