Illuminating the surface of a certain metal alternatrly with light of wavelengths `lambda_(1) = 0.35 mu m` and `lambda_(2) = 0.54 mu m`, it was found that the corresponding maximum velocities of photoelectrons differ by a factor `eta = 2.0`. Find the work function of taht metal.

Explanation

From the last eqution of the previous problem, we find
`eta= ((v_(1))_(max))/((v_(2))_(max)) = sqrt(((1)/(lambda_(0))-(1)/(lambda_(0)))/((1)/(lambda_(2))-(1)/(lambda_(0))))`
Thus `eta^(2)((1)/(lambda_(2))-(1)/(lambda_(0))) = (1)/(lambda_(1)) - (1)/(lambda_(0))`
or `(1)/(lambda_(0)) (eta^(2) - 1) = (eta^(2))/(lambda_(2))-(1)/(lambda_(1))`
and `(1)/(lambda_(0)) = ((eta^(2))/(lambda_(2))-(1)/(lambda_(1)))//(eta^(2)-1)`
So `A = (2picancelh c)/(lambda_(0)) = (2pi cancelh c)/(lambda_(1)) (eta^(2) - (lambda_(2))/(lambda_(1)))/(eta^(2) - 1) = 1.88eV`