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Light of frequency 7.21xx10^(14) Hz is incident on a metal surface .Electrons with a maximum speed of 6.0xx10^(5)m//s are ejected from the surface .What is the threshold frequency for photoemission of electrons? |
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Answer» Solution :Here frequency v`=7.21xx10^(14)` Hz `v_(max)=6.0xx10^(5)m//s` `V_(0)=?h=6.63xx10^(-34)Js` `m=9.1xx10^(-31)Js` `m=9.1xx10^(-31)KG` `implies`From Einstien.s EQUATION of photoelectric effect. `K_(max)=hv-phi_(0)` `therefore (1)/(2)mv^(2)max=hv-hv_(0)` `thereforehv_(0)=hv-(1)/(2)mv^_(max)(2)` `therefore v_(0)=v-(mv_(max)^(2))/(2h)` `=7.21xx10^(14)-(9.1xx10^(-31)xx36xx10^(10))/(2xx6.63xx10^(-34))` `=7.21xx10^(14)-2.47xx10^(14)` `therefore v_(0)4.74xx10^(14) Hz` |
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