A photon of wavelength `5000 lambda` strikes a metal surface with work function `2.20 eV` calculate
aTHe energy of the photon in eV
b. The kinetic energy of the emitted photo electron
c. The velocity of the photoelectron

Explanation

`E = hv= (hc)/(lambda)`
`= ((6.6 xx 10^(-34)) J s(3 xx 10^(8) m s^(-1)))/(5 xx 10^(-3) m) = 3.96 xx 10^(-19) J`
`1 eV = 1.6 xx 10^(-19)J`
`:. E = (3.96 xx 10^(-19) J)/(1.6 xx 10^(-19) J eV^(-2)) = 2.475 eV`
Kinetic energy of the emitted photoelectron work function `= 2.20 eV`
Therefore` KE= 2.475 - 2.20 = 0.275 eV = 4.4 xx 10^(-20) J`
Velocity of the photoelectron
`KE = (1)/(2) mv^(2) = 4.4 xx 10^(-20) J`
Therefore velocity (v)
`sqrt((2 xx 4.4 xx 10^(-20))/(9.1 xx 10^(-31))) = 3.11 xx 10^(5) m s^(-1)`