In a photoelectric effect set-up, a point source of light of power 3.2×10-3W emits monochromatic photon of energy 5eV. The source is located at a distance 0.8 m from the centre of a stationary metallic sphere of work function 3eV and of radius 8×10-3m. The efficiency of photoelectron emission is one for every 106 incident photons. Assume that the sphere is isolated and electrons are instantly swept away after emission. Calculate the number of photoelectrons emitted per second.

Number of photoelectrons emitted per secondn=1106PEπr24πR2 Given, P=3.2×10-3 W, E=5 eV, r=8×10-3 m, R=0.8 m.⇒n=1106×3.2×10-35×1.6×10-19×π×8×10-3×8×10-34π×0.8×0.8⇒n=105sec-1

In a photoelectric effect set-up, a point source of light of power 3.2×10-3W emits monochromatic photon of energy 5eV. The source is located at a distance 0.8 m from the centre of a stationary metallic sphere of work function 3eV and of radius 8×10-3m. The efficiency of photoelectron emission is one for every 106 incident photons. Assume that the sphere is isolated and electrons are instantly swept away after emission. Calculate the number of photoelectrons emitted per second.

Number of photoelectrons emitted per secondn=1106PEπr24πR2 Given, P=3.2×10-3 W, E=5 eV, r=8×10-3 m, R=0.8 m.⇒n=1106×3.2×10-35×1.6×10-19×π×8×10-3×8×10-34π×0.8×0.8⇒n=105sec-1

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

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