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In a photoelectric effect experiment irradiation of a metal with light of frequency `5.2 xx 10^(14) s^(-1)` yields elctrons with maximum kinetic ennergy `1.3 xx 10^(-19) J`.Calculate the threshold frequency `(v_(0))` for the metal |
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Answer» We know that `hv = hv_(0)+ KE` or `v_(0) = v - (KE)/(h)` `KE = 1.3 xx 10^(-19)J, v = 5.2 xx 10^(14)s^(-1), h = 6.626 xx 10^(-34) Js` :. Theshold frequency `v_(0) = 5.2 xx 10^(24)s^(-1) - (1.3 xx 10^(19)J)/(6.626 xx 10^(-34)Js)` `= 5.2 xx 10^(14)s^(-1) - 1.96 xx 10^(14)s^(-1)` `3.24 xx 10^(14) s^(-1)` |
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