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A photosensitive metallic surface has work function hν_0. If photons of energy 2 hν_0 fall on this surface, the electrons come out with a maximum velocity of 4 × 10^6 m/s. When the photon energy is increased to 5 hν_0, then maximum velocity of photoelectrons will be:

1. 2 × 10^7 m/s
2. 2 × 10^6 m/s
3. 8 × 10^6 m/s
4. 8 × 10^5 m/s

Correct answer: 2 × 10^7 m/s

Solution

The kinetic energy of the photoelectrons is given by the photoelectric equation: K.E. = hν - hν_0. For the first case, K.E. = 1/2 m v^2 = hν - hν_0 = h(2ν_0 - ν_0) = hν_0. For the second case, K.E. = h(5ν_0 - ν_0) = 4hν_0. Since K.E. ∝ v^2, the velocity will double, giving v = 2 × 4 × 10^6 = 2 × 10^7 m/s.

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