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) The energy of the photon
Energy (E) `=(hc)/(lambda)=((6.626xx10^(-34)Js)xx(3xx10^(8) ms^(-1)))/((4xx10^(-7)m))=4.97xx10^(-19)J`
`=((1 eV))/((1.602 xx 10^(-19)J))xx(4.97 xx 10^(-19)J)=3.1 eV`
(ii) Kinetic energy of emission
Kinetic energy of emission =E-work function (i.e. kinetic energy of emitted electrons)
`= (3.1-2.13)=0.97 eV`
(iii) Velocity of photoelectron
KE of emission `(1)/(2) mv^(2)=0.97 eV`
`=0.97xx1.602 xx10^(-19)J=0.602 xx 10^(-19)kg m^(2)s^(-2) `
or `" " v^(2)=(2xx0.97xx1.602xx10^(-19)(kg m^(2)s^(-2)))/((9.1xx10^(-31)kg))=0.34 xx10^(12) m^(2)s^(-2)`
or `" " v=(0.34xx10^(12)m^(2)s^(-2))^(1//2)=0.583 xx 10^(6)m s^(-1)=5.83 xx 10^(5) m s^(-1)`.