Find the photoelectric thresholdfor zinc and the maximum velocity of photoelectrons liberated form its surface by electromagnetic radiation with wavelength `250 nm`.

Explanation

The work function of zince is
`A = 3.74 "romane" = 3.74xx1.602xx10^(-19)`Joule
The threshold wavelength for photoelectric effect is given by
`(2pi cancelh c)/(lambda_(0)) = A`
or `lambda_(0) = (2picancelh c)/(A) = 331.6nm`
The maximum velocity of photoelectrons liberated by light of wavelength `lambda` is given by
`(1)/(2)mv_(max)^(2) = 2picancelh c((1)/(lambda)-(1)/(lambda_(0)))`
So `v_(max) = sqrt((4picancelh c)/(m)((1)/(lambda)-(1)/(lambda_(0)))) = 6.55xx10^(5)m//s`