What is the stopping potential, when the metal with work function 0.6 eV is illuminated with light of 2eV

2.6V

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3.6V

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0.8V

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1.4V

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Solution

The correct option is D 1.4V
Given : $ϕ = 0.6 e V$
Energy of incident light $E = 2 e V$
Stopping potential $e V_{o} = E - ϕ$
$∴$ $e V_{o} = 2 - 0.6 = 1.4 e V$
$⟹ V_{o} = 1.4 V$

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