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For a cesium surface ( Work function 1.80 eV) on an average one electron is emitted for every $10^{6}$ photons that reach the surface., Assuming all the ejected electrons took. part in the-charge flow, the current from such a surface if it were illuminated with 600nm light from a 2.00 mW laser is

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$\begin{aligned} 9.68 \times 10^{- 10} A \end{aligned}$

$9.68 \times 10^{- 12} A$

$1.83 \times 10^{- 12} A$

answer is A.

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Detailed Solution

Number of photons incident on metal surface per unit time ,

$n = \frac{p}{h v} = \frac{p λ}{h c} = 6 \times 10^{15}$

Number of photoelectrons liberated per unit time $= 6 \times 10^{9}$

Required current = $6 \times 10^{9} \times 1.6 \times 10^{- 19} = 0.96 nA$

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