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Reviewed CSCA Physics question · Standard

In a photoelectric effect experiment, light of wavelength 250 nm shines on a metal surface. The work function of the metal is 2.5 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Planck constant h = 4.14 × 10⁻¹⁵ eV·s, speed of light c = 3 × 10⁸ m/s)

  1. 2.5 eV
  2. 4.97 eV
  3. 2.47 eV
  4. 7.47 eV
Show the answer and explanation

Correct answer

C. 2.47 eV

Principle or equation

Einstein's photoelectric equation: K_max = hf - φ, where f = c/λ.

Why this answer is correct

Photon energy E = hc/λ = (4.14 × 10⁻¹⁵ eV·s × 3 × 10⁸ m/s) / (250 × 10⁻⁹ m) = 4.97 eV. Then K_max = 4.97 eV - 2.5 eV = 2.47 eV.

Example

If λ = 400 nm and φ = 2.0 eV, then E = 3.10 eV, so K_max = 1.10 eV.

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