Saved Bookmarks
| 1. |
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 ? |
|
Answer» SOLUTION :`phi=2.14eV=2.14xx1.6xx10^(-19)J, upsilon=6xx10^(14)Hz` a. `(1)/(2)mv_("max")^(2)=h upsilon-phi_(0)=6.62xx10^(-34)xx6xx10^(14)-2.14xx1.6xx10^(-19)` `=39.72xx10^(-20)-34.24xx10^(-20)=5.48 xx 10^(-20)=0.548xx10^(-19)J` b. `eV_(0)=(1)/(2)mv_("max")^(2)` `V_(0)=(1)/(2)(m)/(e )v_("max")^(2)=(0.548xx10^(-19))/(1.6xx10^(-19))=0.3425V` C. `(1)/(2)mv_("max")^(2)=0.548xx10^(-19), v_("max")^(2)=(2xx0.548xx10^(-19))/(9.1xx10^(-31))=0.1204xx10^(12)` `v_("max")=344xx10^(3)MS^(-1)` |
|