The photoelectric work function for sodium is 2.7 eV. If light of frequency `9.0xx10^14` Hz falls on sodium, find (a) the stopping potential, (b)the kinetic energy of the most energetic electrons ejected , and (c ) the speed of those electrons.

Explanation

Strategy :We will use energy conservation to find the stopping potential. The most energetic electrons will have energy equal to the stopping potential energy .
The stopping potential energy is equal to the photon's energy minus the work function :
`eV_0=hf-phi`
The stopping potential is then
`V_0=(hf-phi)/e=((9.0xx10^14 Hz)(6.626xx10^(-34) J.s)-(2.7 eV)(1.6xx10^(-19) J//eV))/(1.6xx10^(-19)C)`
The maximum kinetic energy of an electron is then
`K=eV_0=e(1.03 V)=1.03 eV`
and the speed of that electron is
`v=sqrt((2K)/m)=sqrt((2(1.03 eV)(1.6xx10^(-19)J//eV))/((9.11xx10^(-31)kg)))=6.01xx10^5` m/s
The stopping potential is 1.03 V, the maximum kinetic energy of the electrons that are ejected is 1.03 eV, and the speed of the electrons ejected with maximum kinetic energy is `6.01xx10^5` m/s.