Answer
Please see the work below.
Work Step by Step
(a) We know that
$K_{max}=hf-{W_{\circ}}$
We plug in the known values to obtain:
$K_{max}=(6.63\times 10^{-34}J.s)(7.90\times 10^{14}Hz)-(2.24eV)$
This simplifies to:
$K_{max}=3.274eV-2.24eV=1.033eV$
(b) We know that
$f=\frac{W_{\circ}}{h}$
We plug in the known values to obtain:
$f=\frac{2.24eV}{6.63\times 10^{-34}J.s}$
$f=5.41\times 10^{14}Hz$
Thus, the range of frequencies for which no electrons are ejected from the potassium surface is
$4.00\times 10^{14}Hz\le f\le 5.41\times 10^{14}Hz$
