The work function of cesium metal is 2.14 eV .When light of frequency `6 xx 10^(14)` Hz is incident on the metal surface.phtoemission of electrons occurs. What is the :
(a) maximum kinetic energy of the emitted electrons,
(b) stopping potential , and
(c) maximum speed of the emitted photo - electrons ?

Explanation

Here, `phi_(0) = 2.14 eV, upsilon = 6 xx 10^(14) Hz`
(a) `K_(max) = hupsilon-phi_(0) = 6.63 xx10^(-34) xx 6 xx 10^(14 )J -2.14 eV`
`= (6.63 xx 6 xx 10^(-20))/(1.6 xx 106(-19)) eV - 2.14 eV`
`= 2.48 - 2.14 = 0.34 eV`
(b) As `eV_0 = K_(max) = 0.34 eV`
`:.` Stopping potential `V_0 = 0.34eV`
(c) `K_(max) = 1/2 mv_(max)^2 = 0.34 eV`
`= 0 .34 xx 1.6 xx 10^(-19) J`
or `v_(max_^(2) = (2 xx 0.34 xx 1.6 xx 10^(-19))/m`
` = (2xx 0.34 xx 1.6 . 10^(-19))/(9.1 xx 10^(-31)) = 119560 .4 xx10^(6)`
or `v_(max) = 345.8 xx 19^(3) ms^(-1) = 345 .8 kms^(-1)`