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Monochromatic light of wavelength 500 nm falls on a metal whose work function is 2.28 eV. The de Broglie wavelength of the photoelectron emitted is:

1. less than 2.8 × 10⁻⁹ m
2. greater than or equal to 2.8 × 10⁻⁹ m
3. less than or equal to 2.8 × 10⁻¹² m
4. less than 2.8 × 10⁻¹⁰ m

Correct answer: less than 2.8 × 10⁻⁹ m

Solution

Photon energy = 1240/500 = 2.48 eV, so KE = 2.48 - 2.28 = 0.20 eV. de Broglie wavelength = 12.27/sqrt(0.20) A ~ 27.4 A = 2.74e-9 m, which is less than 2.8e-9 m.

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