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Two metallic plates $A$ and $B$ each of area $5 \times 10^{- 4} m^{2}$ are placed parallel to each other at separation of 1 cm. Plate B carries a positive charge. A monochromatic beam of light, with photons of energy $5 e V$ each, starts falling on plate $A$ at $t = 0$ so that $10^{16}$ photons fall on it per square meter per second. Assume that the photoelectron is emitted for every $10^{6}$ incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate $A$ remains constant at the value $2 e V$ .
The number of photoelectrons emitted up to $t = 10 s$ is:

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$5 \times 10^{6}$

$0.5 \times 10^{7}$

$50$

$5 \times 10^{7}$

answer is D.

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Detailed Solution

Area of plates $= 5 \times 10^{- 4} m^{2}$

Number of photoelectrons emitted up to $t = 10 s$

$n = \frac{(no of photons falling in unit area in unit time) \times (area) \times (time)}{10^{6}}$

$\Rightarrow n = \frac{1}{10^{6}} [(10)^{16} \times (5 \times 10^{- 4}) \times (10)]$

$\Rightarrow n = 5 \times 10^{7}$

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