4 eV is the energy of the incident photon and the work function in 2eV. What is the stopping potential

$2 V$

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$4 V$

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$6 V$

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$2 \sqrt{2} V$

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Solution

The correct option is A
$2 V$

$E = W_{0} + e V_{0} \Rightarrow 4 e V + e V_{0} \Rightarrow V_{0} = 2 v o l t$

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