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 the photon `(E) = hv = (hc)/(lamda) = ((6.626 xx 10^(-34) Js) xx(3 xx 10^(8) ms^(-1)))/(4 xx 10^(-7) m) = 4.97 xx 10^(-19) J`
`= (4.97 xx 10^(-19))/(1.602 xx 10^(-19)) eV = 3.10 eV`
(ii) Kinetic energy of emission `((1)/(2) mv^(2)) = hv - hv_(0) = 3.10 - 2.13 = 0.97 eV`
(iii) `(1)/(2) mv^(2) = 0.97 eV = 0.97 xx 1.602 xx 10^(-19) J`
i.e., `(1)/(2) xx (9.11 xx 10^(-31) kg) xx v^(2) = 0.97 xx 1.602 xx 10^(-19) J`
or `v^(2) = 0.341 xx 10^(12) = 34.1 xx 10^(10) or v = 5.84 xx 10^(5) ms^(-1)`