Light of wavelength 3000 Å falls on a metal surface having work function 2.3 eV. Calculate the maximum velocity of ejected electrons.
(Planck's constant , ` h = 6.63 xx 10 ^( - 34 ) ` J.s. , velocity of light ` c = 3 xx10 ^( 8 ) m//s ` , mass of an electron = ` 9.1 xx 10 ^( -31 ) ` kg )

Explanation

Given : ` lamda 3000 Å, v = 3 xx 10 ^( - 7 ) m`,
`phi _ 0 = 2.3 eV = 2.3 xx 1.6 xx 10 ^( -19 ) J = 3.68 xx 10 ^( -19 ) J `
` ( 1 ) /( 2 ) mv _ ( max ) ^ 2 = hv - phi _ 0 `
K.E. ` = ( hc ) / ( lamda ) - phi _ 0 `
K.E. ` = ( 6.63 xx 10 ^( - 34 ) xx 3 xx 10 ^( 8)) /( 3 xx 10 ^( - 7 )) - 3.68 xx 10 ^( - 19 ) `
K.E. = ` 6.63 xx 10 ^( - 19 ) - 3.68 xx 10 ^( - 19 ) `
K.E. = ` ( 6.63 - 3.68 ) xx 10 ^( -19 ) `
K.E. ` = 2.95 xx 10 ^( -19) J `
` therefore ( 1 ) / (2 ) mv _ ( max ) ^ 2 = K.E. `
` therefore v _ ( max ) = sqrt (( 2 K.E.)/( m ) ) `
` = sqrt ( ( 2 xx 2.95 xx 10 ^( - 19)/( 9.1 xx 10 ^( - 31)) ) `
` therefore v _ ( max ) = 8.052 xx 10 ^( 5 ) m// s `