AIPMT PRELIMS 2005 Physics Energy of Photoelectrons MCQ Question
A photosensitive metallic surface has work function, hν₀. If photons of energy 2hν₀ fall on this surface, the electrons come out with a maximum velocity of 4 × 10⁶ m/s. When the photon energy is increased to 5hν₀, then maximum velocity of photoelectrons will be:
16 × 10⁶ m/s
8 × 10⁷ m/s
4 × 10⁵ m/s
8 × 10⁶ m/s
Correct Answer
Detailed Explanation
The kinetic energy of the photoelectrons is given by KE = hν - hν₀. For 2hν₀, KE = hν₀, and for 5hν₀, KE = 4hν₀. Since KE is proportional to v², the velocity is doubled when KE is quadrupled, resulting in 8 × 10⁶ m/s.
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