Find the maximum velocity of photoelectrons emitted by radiation of frequency `3 xx 10^(15)` Hz from a photoelectric surface having a work function of 4.0 eV. Given `h=6.6 xx 10^(-34)` Js and 1 eV = `1.6 xx 10^(-19)` J.
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` (1)/(2) mv_(max)^(2) = hv - phi =6.63 xx 10^(-34) xx 3 xx 10^(15) - 4 xx 1.6 xx 10^(19)` ` v_(max)^(2) = (2[19.89 xx 10^(-19) - 6.4 xx 10^(-19)])/(9.1 xx 10^(-31)) = (26.98 xx 10^(-19))/(9.1xx 10^(31)) = 2.96 xx 10^(12)` ` v_(max) = 1.72 xx 10^(6) m//s ` .
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