InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Light of frequency `7.21xx10^(14)` Hz is incident on 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 ? |
|
Answer» Given, `v =7.21xx10^(14) Hz, m = 9.1xx10^(-31)kg, upsilon_("max")=6xx10^(5)m//s` `KE_("max")=(1)/(2)m upsilon_("max")^(2)= hv-hv_(0)= h(v-v_(0))` `rArr v -v_(0)=((1)/(2)mv_("max")^(2))/(h)=(1)/(2)xx((9.1xx10^(-31))xx(6xx10^(5))xx(6xx10^(5)))/(6.63xx10^(-34)) rArr v -v_(0)=2.47xx10^(14)` `rArr v_(0)= v-2.47xx10^(14)=7.21xx10^(14)-2.47xx10^(14) therefore v_(0) = 4.74xx10^(14)Hz`. |
|