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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be

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 the photoelectric equation: K.E. = hν - φ, where φ is the work function. For the first photon (1 eV), K.E. = 1 - 0.5 = 0.5 eV. For the second photon (2.5 eV), K.E. = 2.5 - 0.5 = 2 eV. The maximum speed is proportional to the square root of K.E., so the ratio of speeds is √(0.5):√(2) = 1:2.

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