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One milliwatt of light of wavelength of 4560 $A^{\circ}$ is incident on a cesium surface. Calculate the electron current liberated. Assume a quantum efficiency of 0.5%.

$\begin {array} {1 1} (a)\;0.184 \: \mu A & \quad (b)\;184 \: \mu Aabc \\ (c)\;Can’t\: be\: determined & \quad (d)\;1.84 \mu A \end {array}$

 

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