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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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-6V

-2.1V

-3.9V

-8.1V

answer is C.

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Detailed Solution

hv = w + KE

KE = hv - W(work function)

= (6 - 2.1)eV

= 3.9 eV

stopping potential must be -3.9 eV

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