Q.

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

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answer is 3.

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Detailed Solution

hv = w + KEKE = hv - W(work function)      = (6 - 2.1)eV      = 3.9 eVstopping potential must be  -3.9 eV
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