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The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be:

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2.4 V

– 2.4 V

1.2 V

– 1.2 V

answer is D.

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Detailed Solution

${eV}_{s} = K_{\max} = \frac{hc}{λ} - φ_{0} \Rightarrow V_{s} = \frac{1}{e} (\frac{h c}{λ} - ϕ_{0}) = \frac{1}{e} (\frac{1240 eV \cdot nm}{200 nm} - 5.01 eV) = \frac{(6.2 - 5.01) eV}{e} = 1.19 V \approx 1.2 V$

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