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A photon having 8 eV energy is incident on metal suraface with threshold frequency of 1.6xx10^(15) Hz .Maximum kinetic energy of photoelectron emitted will be……. h=6.6xx10^(-34)JS,c=3xx10^(8)m//s eV=1.6xx10^(-19)JS |
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Answer» Solution :`(1)/(2)mv_(max)^(2)=E-phi_(0)` `=(E-(hf_(0))/(e))eV [because phi_(0)=(hf_(0))/(e)]` `=(8-(6.6xx10^(-34)xx1.6xx10^(15))/(1.6xx10^(-19)))` |
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