Dual Nature of Matter

JEE Main level, single correct, medium

176 of 200

Question 176 of 200

A metal plate of area 1×10−4m21\times {10}^{-4}{m}^{2} is illuminated by a radiation of intensity 16milliWm216\dfrac{milliW}{{m}^{2}} . The work function of the metal is 5eV.5eV. The energy of the incident photons is 10eV10eV and only 1010% of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: [1eV=1.6×10−19J]\left[1 eV=1.6\times {10}^{-19}J\right]

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Answer: (C) 1011{10}^{11} and 5eV5 eV

Einstein's photoelectric equation
Maximum KE=hCλ−ϕKE=\dfrac{hC}{\lambda }-\phi

=10−5=10-5

=5eV=5eV

Intensity I=(Np)hCλtA⇒Npt=IA(hCλ)I=\dfrac{\left({N}_{p}\right)\dfrac{hC}{\lambda }}{tA}\Rightarrow \dfrac{{N}_{p}}{t}=\dfrac{IA}{\left(\dfrac{hC}{\lambda }\right)}

and number of electrons per second. =(Npt)×10=\left(\dfrac{{N}_{p}}{t}\right)\times 10%

=(IAhCλ)×110=16×10−3×10−410×1.6×10−19×10=\left(\dfrac{IA}{\dfrac{hC}{\lambda }}\right)\times \dfrac{1}{10}=\dfrac{16\times {10}^{-3}\times {10}^{-4}}{10\times 1.6\times {10}^{-19}\times 10}

=1011={10}^{11} per sec

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