Question

The work function for mercury is equivalent to 435kJ/mol photons. What is the kinetic energy, in joules, of the ejected electrons when light of 188 nm strikes a mercury surface?

I know the equation for work function (w) is `E+w=hf` but I don't know where to go from there...
`w=435` kJ

Answer 1

Einstein photo electric equation

`E_k+w=hf`

Where

`E_k->"KE of ejected electron"`

`w->"photo electric work function"=435" kJ/mol"`

`=(435xx1000J/"mol")/(6.023xx10^23mol^-1)=7.22xx10^-19J`

`h->"Planck's constant"=6.62xx10^-34Js`

`f->"frequency of incident light"=c/lambda`
`=(3xx10^8m/s)/(188xx10^-9m)=3/188xx10^17Hz`

where velocity of radiation `c=3xx10^8m/s`
and wave length of radiation `lambda=188nm=188xx10^-9m`

Now

`E_k+w=hf`

`=>E_k=6.62xx10^-34xx3/188xx10^17-7.22xx10^-19J`
`~~10.56xx10^-19-7.22xx10^-19J=3.34xx10^-19J`