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For silver metal `mu_(0)` is `1.13 xx 10^(17)s^(-1)` .What is the maximum energy of the photoelectron produced by shiniong atraviolet light wavelength `1.5` nm on the metal by shining liught wavelength `1.5 nm` on the metal . |
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Answer» `hv = hv_(0) + KE` `KE = hv - hv_(0)` `hc/(lambda) -hv_(0)` `:. KE = ((6.626 xx 10^(-34) Js) xx 3 xx 10^(8) ms^(-1))/(1.5 xx 10^(-9) m)` `-(6.626 xx 10^(-34) Js)(1.13 xx 10^(17) s^(-1))` `= 1.325 xx 10^(-16) J - 7.487 xx 10^(-17) J` `= 0.576 xx 10^(-16) J` |
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