Calculate the number of photons emitted per sec? | TestHub

Passage / Comprehension

In a photoelectric effect setup, a point source light of power

$(\frac{P}{4 π R ^{2}}) = 3.2 \times 1 0^{- 3} W$ emits monoenergetic photons of energy $5.0 e V$ . These photons fall on a metallic sphere of radius $r = 6 mm$ and work function $3.0 e V$ at a distance of $0.6 m$ . The efficiency of photoelectron emission is one for every $1 0^{5}$ incident photons. Assume that the sphere is located and initially neutral, and that photoelectrons are instantly swept away after emission.

;Given hc = 1240 eV - nm]

Calculate the number of photons emitted per sec?

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Answer:

Solution:

Total number of photons per $sec = n$ photons/second

$n \times 5 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 3}$

$n = 4 \times 1 0^{15}$ photons/second

Number of photons falling per second on metallic surface $= x$

$x = n \times \frac{π r ^{2}}{4 π R ^{2}}; r = 0.006 m$ and $R = 0.6 m$

$x = 1 0^{11}$ photons/sec

Number of photoelectrons emitted/sec $= \frac{1 0 ^{11}}{1 0 ^{5}} = 1 0^{6}$

Chemistry - Atomic Structure Question with Solution | TestHub

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JEE Advanced (2028)