A photon iof light with `lambda = 470nm ` falls on a metal surface .As a result photoelectron are ejected with a velocity of `6.4 xx 10^(4) ms^(-1)` .Find
a. The kinetic energy of emited photonelectron
b. The work function (in eV) of the metal surface

Explanation

Kinetic energy of electron `= (1)/(2) mv^(2)`
`rArr KE = (1)/(2)(9.1 xx 10^(-31))(6.1 xx 10^(4))^(2)`
`= 1.86 xx 10^(-23)J`
b. From Einstein's photoelectric equation
`E_(1) = KE + W_(0)`
`rArr W_(0) = E - KE`
`rArr W_(0) = (hc)/(lambda) - KE`
`=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(470 xx 10^(-9)) - 1.86 xx 10^(-21)`
`= 4.21 xx 10^(-19) J`
`rArr W_(0) = 4.21 xx 10^(-19) J = 2.64 eV`