A metal plate area $1 \times 10^{- 4} m^{2}$ is illuminated by a radiation of intensity $16 mW / m^{2}$ the work function of the metal 5ev the energy of the incident photons is 10ev and only 10% of it produces photo electron’s the number of emitted photo electrons per second and their maximum energy respectively will be $[1 eV = 1.6 \times 10^{- 19} J]$

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$10^{14} a n d 10 e V$

$10^{12} a n d 5 e V$

$10^{11} a n d 5 e V$

$10^{10} a n d 5 e V$

answer is C.

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Detailed Solution

Intensity $I = \frac{power}{area} = \frac{p}{A}$

Power $P = \frac{nE}{t}$
Intensity $I = \frac{nE}{At}$

$\begin{array}{l} I = \frac{p}{A} \\ I = 16 \times 10^{- 3} W / m^{2} \\ A = 10^{- 4} m^{2} \\ E = 10 e V = 10 \times 1.6 \times 10^{- 19} J \\ E = 16 \times 10^{- 19} J \\ W = 5 e V \end{array}$

$16 \times 10^{- 3} = \frac{n}{t} \times \frac{16 \times 10^{- 19}}{10^{- 4}}$

$\frac{n}{t} = \frac{16 \times 10^{- 7}}{16 \times 10^{- 19}}$ $\frac{n}{t} = 10^{12}$
So effective number of photo electrons ejected per unit time $= 10^{12} \times 10 %$ $= 10^{12} \times \frac{10}{100}$ $= 10^{11}$
K.E_max $= E - W = (10 - 5) e V$ $= 5 e V$