Answer
$v = 6.76\times 10^5~m/s$
Work Step by Step
We can find the maximum kinetic energy of the ejected electrons:
$E = K_{max}+ \Phi$
$K_{max} = E -\Phi$
$K_{max} = (5.80~eV) -(4.50~eV)$
$K_{max} = 1.30~eV$
$K_{max} = (1.30~eV)(\frac{1.6\times 10^{-19}~J}{1~eV})$
$K_{max} = 2.08\times 10^{-19}~J$
We can find the speed of the fastest electrons that are ejected:
$K_{max} = \frac{1}{2}mv^2 = 2.08\times 10^{-19}~J$
$v^2 = \frac{(2)(2.08\times 10^{-19}~J)}{m}$
$v = \sqrt{\frac{(2)(2.08\times 10^{-19}~J)}{m}}$
$v = \sqrt{\frac{(2)(2.08\times 10^{-19}~J)}{9.109\times 10^{-31}~kg}}$
$v = 6.76\times 10^5~m/s$
