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The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

1. ultra-violet region
2. infra-red region
3. visible region
4. X-ray region

Correct answer: ultra-violet region

Solution

Stopping potential 5 V gives KEmax = 5 eV, so photon energy = 6.2 eV (threshold) + 5 eV = 11.2 eV. Wavelength = 1240/11.2 = 111 nm, which is in the ultra-violet region.

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