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A beam of 10.6 V photons of intensity 2.0W//m^(2)falls on a platinum surface of area 1.0xx10^(-4)m^(2) and work function 5.6eV. If 0.53% of the incident photons eject photo electrons, then find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). |
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Answer» Solution :Number of PHOTONS incident on the metal per second `=("Total energy incident on the metal per second")/("Energy of each photon")` `=("Intensity x AREA")/("Energy of each photon")` Number of electrons emitted per second `=((0.53)/(100))` Number of photons incident on the metal surface Number of electrons emitted per second `=((0.53)/(100))("Intensity x Area")/("Energy of each photon")` `=((2.0)(1.0xx10^(-4)))/((10.6xx1.6xx10^(-19)))XX(0.53)/(100)=6.25xx10^(11)` Minimum KINETIC energy of PHOTO electrons = 0 Maximum kinetic energy is, `K_(max)=E-W_(0)` `=10.6-5.6=5eV`, |
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