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

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$10^{10}$ and $5 e V$

$10^{14}$ and $10 e V$

$10^{12}$ and $5 e V$

$10^{11}$ and $5 e V$

answer is C.

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Detailed Solution

If $n$ photons, each with energy $E$ , are incident on area $A$ per second,

Intensity $I = \frac{n E}{A} \Rightarrow n = \frac{I A}{E} = \frac{(16 \times 10^{- 3}) \times (1 \times 10^{- 4})}{10 \times 1.6 \times 10^{- 19}} = 10^{12} s^{- 1}$

Number of emitted electrons per second is $10 % of n = (\frac{10}{100}) \times 10^{12} = 10^{11} s^{- 1}$

Maximum kinetic energy of emitted photoelectrons $K_{m a x} = E - ϕ_{0} = (10 - 5) eV = 5 eV$ .

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