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

Data: `lambda=3000 Å = 3 xx 10^(-7) m, Phi=2.3 eV = 2.3 xx 1.6 xx 10^(-19) J, h=6.63 xx 10^(-34) J.s, c=3 xx 10^(8) m//s, m=9.1 xx 10^(-31) kg`
`KE_("max") = (hc)/lambda - Phi = (663 xx 10^(-34) xx 3 xx 10^(8))/(3 xx 10^(-7)) -2.3 xx 1.6 xx 10^(-19)`
`=6.63 xx 10^(-19) - 3.68 xx 10^(-19) = 2.95 xx 10^(-19) J`
`KE_("max") = 1/2mv_("max")^(2)`
`therefore` The required velocity, `v_("max") = sqrt((2KE_("max"))/m)= sqrt((2 xx 2.95 xx 10^(-15))/(9.1 xx 10^(-31)))`
`=sqrt(5.9/9.1) xx 10^(6)`
`=0.8054 xx 10^(6) = 8.054 xx 10^(5)` m/s