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Light of two different frequencies, whose photons have energies 2eV and 5eV respectively, successively illuminates a metal of work function 1.0 eV. The ratio of maximum kinetic energy of the emitted photoelectrons will be |
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Answer» `2:5` `(K_(1))/(K_(2))=(E_(1)-phi_(0))/(E_(2)-phi_(0))=((2-1)eV)/((5-1)eV)1:4`. |
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