Illuminating the surface of a certain metal alternatively with light of wavelength `lambda_(1)=0.35 mu m` and `lambda_(2)=0.54 mu m`, it was found that the corresponding maximum velocities of photo electrons have a ratio `eta=12`.
Find the work function of that metal.

Explanation

`(1)/(2)mv_(1)^(2)=(hc)/(lambda_(1))-W`.....(i)
`(1)/(2)mv_(2)^(2)=(hc)/(lambda_(2))W`.....(ii)
Dividing Eq.(i) with Eq.(ii), with `V_(1)=2V_(2)`, we have `((hc)/(lambda_(1))-W)/((hc)/(lambda_(2))-W)`
`3W=4((hc)/(lambda_(2)))-((hc)/(lambda_(1)))=(4xx12400)/(5400)-(12400)/(3500)=5.64eV`
`W=(5.64)/(3)eV=1.88 eV`