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The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

1. 2.4V
2. -1.2V
3. -2.4V
4. 1.2V

Correct answer: 2.4V

Solution

The energy of the incident photon is calculated using E = hc/λ, where h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and λ = 200 nm = 200 × 10⁻⁹ m. Converting the energy to eV and subtracting the work function (5.01 eV), the stopping potential is found to be 2.4 V.

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