In a photoelectric effect set-up, a point source of light of power $3.2 \times 10^{- 3} W$ emits monochromatic photon of energy $5 e V$ . The source is located at a distance 0.8 m from the centre of a stationary metallic sphere of work function $3 e V$ and of radius $8 \times 10^{- 3} m$ . The efficiency of photoelectron emission is one for every $10^{6}$ incident photons. Assume that the sphere is isolated and electrons are instantly swept away after emission. Calculate the number of photoelectrons emitted per second.

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$10^{- 4} \sec^{- 1}$

$10^{4} \sec^{- 1}$

$10^{- 5} \sec^{- 1}$

$10^{5} \sec^{- 1}$

answer is C.

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Detailed Solution

Number of photoelectrons emitted per second

$n = (\frac{1}{10^{6}}) (\frac{P}{E}) (\frac{π r^{2}}{4 π R^{2}})$

Given, $P = 3.2 \times 10^{- 3} W$ , $E = 5 eV$ , $r = 8 \times 10^{- 3} m$ , $R = 0.8 m$ .

$\Rightarrow n = (\frac{1}{10^{6}}) \times (\frac{3.2 \times 10^{- 3}}{5 \times 1.6 \times 10^{- 19}}) \times (\frac{π \times 8 \times 10^{- 3} \times 8 \times 10^{- 3}}{4 π \times 0.8 \times 0.8})$

$\Rightarrow n = 10^{5} \sec^{- 1}$

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