A photoelectric surface has work function 2eV. What is the maximum velocity of the photoelectrons ejected by light of wavelength `3000Å`
(Take, `h = 6 . 6 xx 10 ^(-34) J-s e, = 1.6 xx 10^(-19) C, m_(e) = 9 xx 10 ^(-31) kg, c = 3 xx 10^(8) m//s`)

Explanation

Given, work function , `phi = 2 eV = 2 xx 1.6 xx 10 ^(-19) J`
`= 3 . 2 xx 10^(-19) J`
Wavelength, `lambda = 3000 Å`
According to Einstein.s photoelectric equation
`KE_("max")= hv - phi_(0) rArr (1)/(2) mv_("max")^(2) = (hc)/( lambda) - phi_(0)`
`rArr (1)/(2) xx 9 xx 10 ^(-31) xx v_("max")^(2)`
` = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(3000 xx 10^(-10)) - 3.2 xx 10 ^(-19)`
`rArr 4.5 xx 10 ^(-31) xx v_("max")^(2) =(19.8 xx 10^(-26))/(3xx10^(-7))-3.2 xx 10^(-19)`
`rArr 4.5 xx 10 ^(-31) xx v_("max")^(2) = 6.6xx 10^(-19) -3.2 xx 10^(-19)`
`rArr 4.5 xx 10^(-31) xx v_("max")^(2) = 3.4 xx 10^(-19)`
`rArr v_("max")^(2) = (3.4 xx 10^(-19))/(4.5 xx 10^(-31))`
`rArr v_("max")^(2) = 0.75 xx 10^(12) rArr v_("max") = sqrt(0.75 xx 10^(12))`
`:. v_("max") = 0.866 xx 10^(6) m//s`