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

` E//"photon" = ( hc)/ ( lambda) = ( 6.626 xx 10^(-34) xx 3.0 xx 10^8)/( 4 xx 10(-7))`
` = 4. 97 xx 10 ^(-19) J`
(b) ` E=1/2 m u^2 + w`
` 4. 97 xx 10^(-19) = 1/2 m u^2 + 2.13 xx 1. 6022 xx 10^(-19)`
`therefore k.E. = 1/2 m u^2 = 4. 97 xx 10^(-19) -3. 41 xx 10^(-19)`
` = 1. 56 xx 10^(-19) J = 0. 97 eV`
( c) ` 1/2 m u^2 = K.E.`
` 1/2 xx 9. 108 xx 10^(-31) xxu^2 = 1. 56 xx 10^(-19)`
` therefore u = 5. 85 xx 10^5 ms^(-1)`.