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Light of wavelength 5000\AA and intensity 39.8 W m^{-2} is incident on a metal surface. If only 1% photons of incident light emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be nearly -

Option: 1

10^{18}


Option: 2

10^{22}


Option: 3

10^{24}


Option: 4

10^{26}


Answers (1)

best_answer

\begin{aligned} \mathrm{n}_{\mathrm{Ph}} & =5 \times 10^{24} \mathrm{IA} \lambda \\ & =5 \times 10^{24}\left(\frac{1}{100} \times 39.8\right) \times 1 \times 5000 \times 10^{-10} / \mathrm{sec} \end{aligned}

Posted by

Sanket Gandhi

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