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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 which relation is correct?

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

This option correctly applies the photoelectric effect principles, where the kinetic energy of emitted electrons is proportional to the difference in photon energy, which is determined by the frequencies of the incident light. The relationship between the speeds of the emitted electrons and the frequencies of the light is derived from the equation for kinetic energy, leading to the correct form of the difference of squares.

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