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In a photoelectric effect set up, a poin...

In a photoelectric effect set up, a point source of light of power `3.2 xx10^(-3) W` emits monochromatic photons of energy `5eV`.The source is located at a distance of a stationary metallic sphere of work function `3eV` and radius `8xx10^(-3)m`.The efficiency of photoelectron emission is one for every `10^(6)` incident photons.Assume that the sphere is isolated and initially neutral and the photoelectrons are initially swept away after emission.
Find the number of photons emitted per second

Text Solution

Verified by Experts

Energy of a single phton `E = 5.0 eV = 8 xx 10^(-19) J`
Power of source `P = 3.2 xx 10^(-3) W`
`:.` number of photons emitted per second `n = (P)/(E) = (3.2 xx 10^(3))/(8 xx 10^(-19)) = 4 xx 10^(15)//s`
The number of photons incident per second on metal surface is `n_(0) = (n)/(4piR^(2)) xx pir^(2)`
`n_(0) = (4 xx 10^(15))/(4pi(0.8)^(2)) xx pi(8 xx 10^(-3)) = 1.0 xx 10^(11) "photon"//s`

Number of electrons emitted `= (1.0 xx 10^(11))/(10^(6)) = 10^(5)//s`
`KE_(max) = hv - phi= 5.0 -3.0 = 2.0 eV`
The photonelectron emission stops, when the metalli sphere acquires stopping potential
As `KE_(max) - 2.0 eV rArr "Stoppng potential" V_(0) = 2V rArr 2 = (q)/(4piepsi_(0)r) rArr q = 1.78 xx 10^(12) C`
Now charge `q = ("number of electron"//"second") xx l xx e rArr t = (1.78 xx 10^(-12))/(10^(5) xx 1.6 xx 10^(-19)) = 111s`
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Knowledge Check

  • When metal surface is incident with photon after how much time photoelectron are emitted?

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