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Two identical photocathodes are illuminated by light of frequencies f1 and f2. If the emitted photoelectrons have speeds v1 and v2 respectively, then

1. v1² − v2² = (2h/m)(f1 − f2)
2. v1 + v2 = [2h/m (f1 + f2)]1/2
3. v1² + v2² = (2h/m)(f1 + f2)
4. v1 − v2 = [2h/m (f1 − f2)]1/2

Correct answer: v1² − v2² = (2h/m)(f1 − f2)

Solution

(1/2)m v^2 = h f - phi for each: v1^2 = (2/m)(h f1 - phi), v2^2 = (2/m)(h f2 - phi). Subtracting cancels phi: v1^2 - v2^2 = (2h/m)(f1 - f2).

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