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Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is:

1. 1:4
2. 1:2
3. 1:1
4. 1:5

Correct answer: 1:2

Solution

The maximum kinetic energy of the emitted electrons is given by K.E. = E - W, where E is the photon energy and W is the work function. For 1 eV photon, K.E. = 1 - 0.5 = 0.5 eV. For 2.5 eV photon, K.E. = 2.5 - 0.5 = 2 eV. Since speed is proportional to the square root of kinetic energy, the ratio of speeds is √(0.5):√(2) = 1:2.

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