Light of wavelength 400 nm shines on a metal surface with work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Planck's constant h = 4.14 × 10⁻¹⁵ eV·s, speed of light c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)
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Correct answer
B. 1.1 eV
Principle or equation
Einstein's photoelectric equation: K_max = hf - φ, where f = c/λ.
Why this answer is correct
f = c/λ = (3×10⁸)/(400×10⁻⁹) = 7.5×10¹⁴ Hz. hf = (4.14×10⁻¹⁵ eV·s)(7.5×10¹⁴ Hz) = 3.105 eV. K_max = 3.105 - 2.0 = 1.105 eV ≈ 1.1 eV.
Example
If λ = 500 nm and φ = 2.0 eV, hf = 2.48 eV, K_max = 0.48 eV.
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