Photon of wavelength `4xx 10^(7)` strikes on metal surface the work function of the metal being 2.13 eV Calculate (i) The energy of the photon ev (ii) The kinetic energy of the emission and (iii) the velocity of the photoelectron .
`(1 eV = 1.6020 xx 10^(19) J ) `

Explanation

1.9 eV = `1.9 xx 1.602 xx 10^(19) J`
`E_(0) = 3.0438 xx 10^(19) J`
(a ) Threshold wavelength
`lambda_(0) = (hc )/( E_(0))`
`=((6.626 xx 10^(34) js)(3.0 xx 10^(8) ms^(-1)))/( (1.9xx 1.602 xx 10^(19)J))`
(b ) Threshold frequency (v ) :
`v_(0) = (E_(0))/( h )`
`=0.45937 xx 10^(15) s^(-1) =4.594 xx 10^(14) s^(1)`
kinetic energy (kE)
`KE= (1)/(2) mV^(2) = (E- E_(0))`
`=0.9322 xx 10^(19) J`
`= 9.322 xx 10^(20)J`
Velocity of photoelectron ( v) :
where KE `= (1)/(2) mv^(2)`
Velocity ( v) =`((KE xx 2)/( m )) = ((9.322 xx 10^(20) J xx 2)/( 9.1 xx 10^(31) kg )) ^(1)/(2) `
` = 2.0488 xx 10^(11) (1)/(2) (j Kg^(-1))`
`4.53 xx 10^(5) ms^(-1) J = kg ms^(-1) J`