The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will be

Options

A. 10 m/s
B. `1xx10^3 m//s`
C. `1xx10^4 m//s`
D. `1xx10^6 m//s`

Answer

D

Explanation

`K_("max") = hf-phi_0`
`implies 1/2mv_("max")^2 = (12375)/lamda -phi_0`
`implies 1/2mv_("max")^2 = ((12375)/3000-1)eV=3.125eV`
`v_("max")= sqrt((2xx3.125xx 1.6 xx10^(-19))/(9.1 xx10^(-31)))=10^6m/s`