A metal plate of area $1 \times 10^{- 4} m^{2}$ is illuminated by a radiation of intensity $16 \frac{mW}{m^{2}}$ . 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 $[1 eV = 1.6 \times 10^{- 19} J]$

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

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

$10^{14}$ and $5 e V$

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

answer is C.

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Detailed Solution

$ϕ = 5 eV, E = 10 eV ∴ {KE}_{\max} = E - ϕ = 10 - 5 = 5 eV$

Energy incident on the plate,

$E' = (16 \times 10^{- 3}) \times (1 \times 10^{- 4}) \frac{J}{s} = 16 \times 10^{- 7} \frac{J}{s} = 10^{13} \frac{eV}{s}$

Number of photo electrons emitted per second

$= 10 %$ of $\frac{E'}{E} = 0.1 \times \frac{10^{13} \frac{e V}{s}}{10 e V} = 10^{11} s^{- 1}$

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