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Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The wavelength of the emitted electron is

1. < 2.8 × 10⁻⁹ m
2. > 2.8 × 10⁻⁹ m
3. < 2.8 × 10⁻¹² m
4. > 2.8 × 10⁻¹² m

Correct answer: > 2.8 × 10⁻⁹ m

Solution

The energy of the incident photon is calculated using E = hc/λ. Subtracting the work function gives the kinetic energy of the emitted electron. Using de Broglie's equation, λ = h/p, the wavelength of the electron is found to be greater than 2.8 × 10⁻⁹ m.

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