A metallic surface is illuminated alternatively with light of wavelenghts `3000 Å` and `6000 Å`. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1 . Calculate the work function of the metal and the maximum speed of the photoelectrons in two cases.

Answer

`1.81 eV`

Explanation

Maximum kinetic energy of photo electrons
`K_(max)=1/2 mv_(max)^(2)=(hc)/lambda-phi_(0)`
Now let `3000 Å=lambda` then `6000 Å=2lambda`
`:. ((hc)/lambda-phi_(0))/((hc)/(2lambda)-phi_(0))=((v_(max))_(1)^(2))/((v_(max))_(2)^(2))=9/1`
`implies phi_(0)=(7hc)/(16 lambda)=(7xx6.62xx10^(-34)xx3xx10^(8))/(16xx3000xx10^(-10)xx1.6xx10^(-19))=1.81 eV`