Calculate the velocity of a photo-electron if the work function of the target material is `1.24eV` and the wavelength of incident light is `4.36xx10^(-7)`m. What retarding potential is necessary to stop the emission of the electrons?

Explanation

As `KE_(max) = hv + phi rArr 1/2 mv_(max)^(2) = hv - phi = (hc)/(lambda) - phi`
`v_(max) = sqrt((2(sqrt(hc)/(lambda)-phi))/(m)) = sqrt(2(((6.63 xx 10^(-34) xx 3 xx 10^(8))/(4.36 xx 10^(-7)) - 1.24 xx 1.6 xx 10^(19)))/(9.11 xx 10^(-31))) = 7.523 xx 10^(5) m//s`
`:.` The speed of a photoelectron acan be any value between `0` and `7.43 xx 10^(5) m//s`
If `V_(0)` is the stopping potential, then `eV_(0) = 1/2 mv_(max)^(2)`
`rArr V_(0) = 1/2 (mv_(max)^(2))/(e) = (hc)/(elambda) - phi/e = (2400)/(4360) - 1.24`
`= 1.60 V [:' (hc)/(e) = 12400 xx 10^(-10)V-m]`