Light of wavelength `5000Å` falls on a sensitive plate with photoelectric work function of 1.9 eV. The K.E. of the photo electron emitted will be

Explanation

Here `lambda = 5000 Å = 5 xx 10^(-7)m`
`W_(0) = 1.9 eV`
(i) Energy of a photon,
`E = (hc)/(lambda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(5 xx 10^(-7))J = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(5 xx 10^(-7) xx 1.6 xx 10^(-19))eV`
E = 2.475 eV
(ii) K.E of a photoelectron,
`K.E = h upsilon - W_(0) = 2.475 - 1.9 = 0.575 eV`
(iii) Let `V_(0)` be the stopping potential. Then
`eV_(0) = (1)/(2) mv^(2) =` K.E of a photoelectron
`V_(0) = (0.575)/(e) eV`
`V_(0) = 0.575 V`