Answer
$K_{max} = 10.1~eV$
Work Step by Step
We can find the maximum kinetic energy of the ejected electrons:
$E = hf = K_{max}+ \Phi$
$K_{max} = hf -\Phi$
$K_{max} = (6.63\times 10^{-34}~J\cdot s)(3.0\times 10^{15}~Hz) -(2.3~eV)$
$K_{max} = (1.989\times 10^{-18}~J) -(2.3~eV)$
$K_{max} = (1.989\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J}) -(2.3~eV)$
$K_{max} = 12.4~eV -2.3~eV$
$K_{max} = 10.1~eV$
