Physics, asked by srikalaravipati1406, 8 months ago

a point source of light emits photon of energy 4.8eV at the rate of 10^5​

Answers

Answered by adarshedu88
0

Answer:

Explanation: In a photoelectric effect set-up a point source of light of power 3.2×10  

−3

 W emits monoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0×10  

−3

m. The efficiency of photoelectron emission is one-for every 10  

6

incident photons. Assume that the sphere is isolated and initially natural and that photoelectron are instantly swept away after emission.

Calculator the number of photoelectron emitted per second.

Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectron emitted.

It is observed that the photoelectron emission stop at a certain time t after the light source is switch on. Why?

Evaluate the time t.

Answered by bhuvankumarapatripow
0

Answer:

ANSWER

Energy of emitted photons 

E1=0.5 eV=8×10−19

Energy emitted per second by the source

E2=3.2×10−3 J

No. of photons emitted per second =E1E2=4×1015

No. of photons incident on the sphere per second

n=4π(0.8)24×1015×π(8×10−3)2≈1011 per second

No, of photoelectron emitted per second 

n′=1061

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