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A metal has a work function of 3.5 eV in the photoelectric effect. If the emitted photoelectrons are just stopped by a retarding potential of -1.2 V, then which statement is correct?

1. The incident photon energy is 4.7 eV.
2. The incident photon energy is 2.3 eV.
3. Using light of higher frequency will increase the photoelectric current.
4. For photons of energy 3.5 eV, the photoelectric current will be maximum.

Correct answer: The incident photon energy is 4.7 eV.

Solution

KEmax = eVs = 1.2 eV. Incident photon energy = phi + KEmax = 3.5 + 1.2 = 4.7 eV. (Photocurrent depends on intensity, not frequency, so the other statements are false.)

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