$W_o = 2.28$ eV

$KE_{max} = 2.00$ eV

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(2.00 text{ eV} + 2.28 text{ eV}) times left(dfrac{1.6 times 10^{-19} text{ eV}}{1 text{ eV}}right)}{6.626 times 10^{-34} text{ Js}} \
&= boxed{1.03 times 10^{15} text{ Hz}}
end{align*}
$$

$W_o = 4.58$ eV (as seen from Example 24.5)

$KE_{max} = 1.00$ eV

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(1.00 text{ eV} + 4.58 text{ eV}) times left(dfrac{1.6 times 10^{-19} text{ eV}}{1 text{ eV}}right)}{6.626 times 10^{-34} text{ Js}} \
&= boxed{1.35 times 10^{15} text{ Hz}}
end{align*}
$$

$W_o = 2.5 text{ eV} times left(dfrac{1.6 times 10^{-19} text{ J}}{1 text{ eV}}right) = 4.0 times 10^{-19}$ J

$KE_{max} = 1.8 times 10^{-19} text{ J}$

$textbf{Find: } f$

$textbf{Formula: } KE_{max} = hf – W_o$

Using this equation, we can obtain $f$ as follows:

$$
begin{align*}
KE_{max} &= hf – W_o \
f &= dfrac{KE_{max} + W_o}{h} \
&= dfrac{(1.8 times 10^{-19} text{ J})+ (4.0 times 10^{-19} text{ J})} {6.626 times 10^{-34} text{ Js}} \
&= boxed{8.8 times 10^{14} text{ Hz}}
end{align*}
$$