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)`
`:. V_(0)=W_(0)/h`
`=(1.9xx1.602xx10^(-19))/(6.626xx10^(-24)Js)(leV=1.602xx10^(-19)J)`
`=4.59xx10^(14) s^(-1)`
b. `lambda_(0)=c/v_(0)=(3.0xx10^(8) ms^(-1))/(4.59xx10^(14)s^(-1))=6.54xx10^(-7) m`
`=654xx10^(-9)m=654 nm`
c. `KE` of ejected electron `=h(v-v_(0))`
`=hc(i/lambda-1/lambda_(0))`
`=(6.626xx10^(-34)Js)(3.0xx10^(8) m s^(-1))xx(1/(500xx10^(-9)m)-1/(654xx10^(-9)m))`
`=(6.626xx3.0xx10^(-26))/10^(-9)(154/(500xx654))J=9.36xx10^(-20)J`
`KE=1/2mv^(2)=9.36xx10^(-20)J or kg m^(2) s^(-2)`
`:. 1/2xx(9.11xx10^(-31)kg)v^(2)=9.36xx10^(-20) kgm^(2)s^(-2)`
or `v^(2)=2.055xx10^(11) m^(2) s^(-2)`
`=20.55xx10^(10) m^(2) s^(-2)`
or `v=4.53xx10^(5) m s^(-1)`