Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?

-1.9V

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-5V

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-3.9V

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-8.1V

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Solution

The correct option is C
-3.9V

K_max= eV₀= E – Φ₀= 6 – 2.1 = 3.9 eV

Therefore, stopping potential V₀= 3.9 V

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