A photon of wavelength ` 4 xx 10^(-7) m` strikes on metal surface , the work function fo the metal being ` 2. 13 eV` Calculate :
(i) the energy of the photon (ev)
(ii) the kinetic energy fo the emission and
the velocity fo the photoelectron `( 1 eV = 1 , 6020 xx 10^(-19) J)`,

Explanation

(i) Energy of photon
`E = hv = (hc)/(lambda)`
`lambda = 4.0 xx 10^(-7) m, " " c = 3.0 xx 10^(8) ms^(-1)`
`h = 6.626 xx 10^(-34) Js`,
`:. E = ((6.626 xx 10^(-34)Js) xx (3.0 xx 10^(8)m))/((4.0 xx 10^(-7)m))`
`= 4.97 xx 10^(-19)J`
or `" " = (4.97 xx 10^(-19))/(1.6020 xx 10^(-19)) = 3.10 eV`.
(ii) Kinetic energy of emission
K.E. = `hv - hv_0`
`hv = 3.10 eV and hv ("work function") = 2.13 eV`
`:. K .E. = 3.10 - 2.13 = 0.97 eV`
(iii) Now, `K.E. = 1/2 mv^2`
or `1/2 mv^2 = 0.97 eV = 0.97 xx 1.602 xx 10^(-19) J`
or `1/2 xx (9.11 xx 10^(-31) kg) v^2 = 0.97 xx 1.602 xx 10^(-19) J`
`v^2 = (2 xx 0.97 xx 1.602 xx 10^(-19) kg m^2 s^(-2))/(9.11 xx 10^(-31) kg)`
`=34.12 xx 10^(10) m^2 s^(-2)`
`:. v = 5.84 xx 10^(5) ms^(-1)`