Photo-energy 6 eV are incident on a surface of work function 2.1 eV. What is the stopping potential

-5 V

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-1.9 V

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-3.9 V

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-8.1 V

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Solution

The correct option is C -3.9 V
By using Einstein's equation $E = W_{0} + K_{m a x} \Rightarrow 6 = 2.1 + K_{m a x} \Rightarrow K_{m a x} = 3.9 e V$
Also $V_{0} = - \frac{K_{m a x}}{e} = - 3.9 V$ .

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