A metal plate of area \(1\times10^{-4}\, m^2\) is illuminated by a radiation of intensity \(16\, m\,W/m^2\). The work function of the metal is \(5\,eV\). The energy of the incident photons is \(10\, eV\) and only \(10\%\) of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : \([1\, eV = 1.6\times10^{-19}\, J]\)

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Correct Answer

(C) \(10^{11}\) and \(5\, eV\)

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Detailed explanation

\(n = \frac{{16 \times {{10}^{ - 3}} \times {{10}^{ - 4}}}}{{10 \times 10 \times {{10}^{ - 19}}}}\) \( \simeq 1.6 \times {10^{11}}\)

\({({\text{K}} \cdot {\text{E}})_{\max }} = (10 - 5)\,{\text{eV}} = 5\,{\text{eV}}\)

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