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A metal has threshold frequency f0. When light of frequency 2*f0 is incident on the metal, the maximum velocity of emitted photoelectrons is v1. When light of frequency 5*f0 is incident, the maximum velocity of emitted photoelectrons is v2. Find the ratio v1: v2.
- 1: 4
- 1: 2
- 2: 1
- None of these
Correct answer: 1: 2
Solution
Einstein photoelectric equation: (1/2)m*v² = h*(f - f0). For f = 2f0: (1/2)m*v1² = h*(2f0 - f0) = h*f0 For f = 5f0: (1/2)m*v2² = h*(5f0 - f0) = 4h*f0 Dividing: v1²/v2² = h*f0 / (4h*f0) = 1/4 v1/v2 = 1/2 Ratio v1: v2 = 1: 2.
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