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Light of wavelength 5000 Å falls on a sensitive plate with photo-electric work function of 1.9 eV. The kinetic energy of the photo-electrons emitted will be

1. 0.58 eV
2. 2.48 eV
3. 1.24 eV
4. 1.16 eV

Correct answer: 0.58 eV

Solution

The energy of the incident photon is calculated using E = hc/λ. Substituting h = 4.1357 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m/s, and λ = 5000 Å = 5 × 10⁻⁷ m, we get E = 2.48 eV. The kinetic energy of the photoelectrons is then KE = E - work function = 2.48 eV - 1.9 eV = 0.58 eV.

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