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The work function of caesium metal is 2.14 eV.When light of frequency 6xx10^(14) Hz is incident on the metal surface ,photoemission of electrons occurs.What is the (a)Maximum kinetic energy of the emitted electrons. (b)Stopping potential ,and (c )Maximum speed of the emitted photoelectrons? |
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Answer» Solution :Work function `phi_(0)`=2.14 eV Frequency of light `v_(0)=6xx10^(14)Hz` `h=6.63xx10^(-34)` JS, l eV=`1.6xx10^(-19)J` (a) Photoelectric equation , `K_(max)=hv-phi_(0)` `=((6.63xx10^(-34)xx6xx10^(14))/(1.6xx10^(-19))-2.14)eV` =(2.48625-2.14) eV (B) `K_(max)=eV_(0)` `therefore V_(0)=(K_(max))/(e)=(0.34eV)/(e) therefore V_(0)=0.34 V` (c )`K_(max)=0.34eV` `therefore (1)/(2) mv_(max)^(2)=0.34xx1.6xx10^(-19)` `therefore V_(max)^(2)=(2xx0.346xx1.6xx10^(-19))/(m)` `=(2xx0.34xx1.6xx10^(-19))/(9.1xx10^(-31))` `therefore v_(max)=sqrt((0.692xx1.6xx10^(-19))/(9.1xx10^(-31)))` `therefore v_(max)=sqrt(0.12167xx10^(12))` `=0.345775xx10^(6)~~345.8xx10^(3)m//s` `therefore v_(max)=345.8 km//s` Note:Textbook answer is DIFFERENT. |
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