Light falling on metallic surface of work function 2eV is given by B=100[sin(5×10^15)t +cos(8×10^15)t]. Then maximum kinetic energy of the photoelectrons is______

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Sriram · Answer

Omega=5×10^15 and B0=100 As kmax=homega-2π/2π (by applying v=omega/2π) On substituting the values we get 3.3ev

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Light falling on metallic surface of work function 2eV is given by B=100[sin(5×10^15)t +cos(8×10^15)t]. Then maximum kinetic energy of the photoelectrons is______

Light falling on metallic surface of work function 2eV is given by B=100[sin(5×10^15)t +cos(8×10^15)t]. Then maximum kinetic energy of the photoelectrons is______

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Light falling on metallic surface of work function 2eV is given by B=100[sin(5×10^15)t +cos(8×10^15)t]. Then maximum kinetic energy of the photoelectrons is______

Light falling on metallic surface of work function 2eV is given by B=100[sin(5×10^15)t +cos(8×10^15)t]. Then maximum kinetic energy of the photoelectrons is______

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