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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 `v=7.21xx10^(14)Hz and v_(max)=6.0xx10^(5)m//s` Using the relation`H(v-v_(0))=(1)/(2)mv_(max)^(2),` we have `v-v_(0)=(mv_(max)^(2))/(2h)=(9.11xx10^(-31)xx(6.0xx10^(5))^(2))/(2xx6.63xx10^(-34))=2.48xx10^(14)` `v_(0)=v-2.48xx10^(14)=7.21xx10^(14)-2.48xx10^(14)=4.73xx10^(14)Hz`. |
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