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When a beam of 10.6 eV photons of intens...

When a beam of 10.6 eV photons of intensity 2.0 `W//m^(2)` falls on a platinum surface of area `1.0xx10^(-4)m^(2)` and photons eject photoelectrons. Find the number of photoelectrons emitted per second

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Energy incident on the surface per second
`P=IA`
`=2.0xx1.0xx10^(-4)`
`=2xx10^(-4)Js^(-1)`
Energy of each photon
`=10.6eV`
`=10.6xx1.6xx10^(-19)J`
Number of photons incident on the surface
`=(2xx10^(-4))/(10.6xx1.6xx10^(-19))`
Number of photoelectrons emitted
`=(0.53)/(100)x(2xx10^(-4))/(10.6xx1.6xx10^(-19))=6.25xx10^(11)`
According to Einstein's photoelectric equation, maximum KE of photoelectrons,
`E_k=epsi - W=10.6eV-5.6eV`
`=5eV`
Minimum kinetic energy of photoelectrons = zero
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