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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 work function `5.6 eV, 0.53%` of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energy (in eV).
Take `1 eV = 1.6xx 10^(-19) J`.

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when a beam of 10.6 eV photons of intensity 2.0 W//m^(2) falls on a platinum surface of area 1.0 xx 10^(4) m^(2) and work function 5.6 eV , 0.53 % of the incidentphotons eject photoelectrons find the number of photoelectrons emited per second and their minimum energies (in eV)Take 1 eV= 1.6 xx 10^(-19) J

A metal plate of area 1 xx 10^(-4) m^(2) is illuminated by a radiation of intensity 16m W//m^(2) . The work function of the metal is 5eV. The energy of the incident photons is 10eV. The energy of the incident photons is 10eV and 10% of it produces photo electrons. The number of emitted photo electrons per second their maximum energy, respectively, will be : [1 eV = 1.6 xx 10^(-19)J]

Knowledge Check

  • A light of intensity 16 mW and energy of each photon 10 eV incident on a metal plate of work function 5 eV and area 10^(-4)m^(2) then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is 10% .

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    B
    10 eV, `10^(12)`
    C
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  • Ultraviolet light of wavelength 66.26 nm and intensity 2 W//m^(2) falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons, and surface area of metal surface is 4 m^(2) , how many electrons are emitted per second?

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  • in the question number 63, the frequency of emitted photon due to the given transition is (h=6.64xx10^(-34)Js,1eV=1.6xx10^(-19)J)

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