The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron

Explanation

(a) Work function `(W_(0))=hv_(0)" where "v_(0)` is the thereshold frequency.
`v_(0)=W_(0)/h=(1.9 xx 1.6021 xx 10^(-19))/(6.626 xx 10^(-34)`
`(1eV=1.6021 xx 10^(-19)J)`
`=4.594 xx 10^(14) s^(-1)`
(b) Threshold wavelength `(lambda_(0)` is given by
`lambda_(0)=c/v_(0)`
`=(3.0 xx 10^(8))/(4.594 xx 10^(14))`
` =6.53 xx 10^(-7)m=653nm`
(c) `v=c/lambda=(3.0 xx 10^(8))/(500 xx 10^(-9))=6.0 xx 10^(14) s^(-1)`
Kinetic energy of the ejected electron
`=h(v-v_(0))`
`=6.626 xx 10^(-34) (6.0 xx 10^(14) -4.594 xx 10^(14))`
`=9.32 xx 10^(-20)J`
Kinetic energy `=1/2 mv^2`
Velocity of ejected photoelectron `=((2 xx K.E.)/(m))^(1//2)`
`=(2 xx 9.32 xx 10^(-20))/(9.11 xx 10^(-31))^(1//2)`
`=4.523 xx 10^(5)ms^(-1)`.