1.

What a certain was metal was irradiatiobn with light of frequency `1.6 xx 10^(16) Hz` the photoelectron emitted but the kinetic energy as the photoelectron emitted when the same metal was irradiation with light of frequency `1.0 xx 10^(16) Hz` .Calculate the threslold frequency `(v_(0))` for the metal

Answer» `hv = hv_(0) = KE`
`KE_(1) - h(v - v_(0))`
`KE_(2) = h(v_(2) - v_(0)) = (KE_(1))/(2)`
`:. (v_(2) - v_(v_(0)))/(v_(1) - v_(0)) = (1)/(2) rArr (1.0 xx 10^(16) - v_(0))/(1.6 xx 10^(16) - v_(0))m = (1)/(2)`
`rArr v_(0) = 4 xx 10^(15)Hz`


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