Answer
See the detailed answer below.
Work Step by Step
$\Rightarrow$ We know that the threshold frequency is given by
$$f_0=\dfrac{E_0}{h}\tag a$$
$\Rightarrow$We also know that the corresponding threshold wavelength is given by
$$\lambda_0=\dfrac{c}{f_0}$$
Plugging from (1);
$$\lambda_0=\dfrac{hc}{E_0}\tag b$$
$\Rightarrow$ We know that the maximum kinetic energy of the photoelectrons is given by
$$K_{\rm max}=\frac{1}{2}mv_{\rm max}^2=hf-E_0$$
Thus,
$$v_{\rm max}=\sqrt{\dfrac{2(hf-E_0)}{m}}$$
where $f=c/\lambda$;
$$v_{\rm max}=\sqrt{\dfrac{2\left(\dfrac{hc}{\lambda}-E_0\right)}{m}}\tag c$$
$\Rightarrow$ Now we need to recall the stopping potential which is given by
$$V_{\rm stop}=\dfrac{K_{\rm max}}{e}$$
$$V_{\rm stop}=\dfrac{\dfrac{hc}{\lambda}-E_0}{e}\tag d$$
Now we need to use these 4 formulas from (1) to (4) to find the values of these variables for both the potassium and gold.
We will use the data in Table 38.1; but don't forget to convert $E_0$ for eV to J to find the right answer, as we did below.
$$\color{blue}{\bf [a]}$$
$\;\underline{\underline{\textbf{For Potassium:}}}$
$$f_0=\dfrac{E_0}{h}=\dfrac{(2.3\times 1.602\times 10^{-19})}{(6.626\times 10^{-34})}=\color{red}{\bf 5.56\times 10^{14}}\;\rm Hz$$
$\;\underline{\underline{\textbf{For Gold:}}}$
$$f_0=\dfrac{E_0}{h}=\dfrac{(5.1\times 1.602\times 10^{-19})}{(6.626\times 10^{-34})}=\color{red}{\bf 1.23\times 10^{15}}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
$\;\underline{\underline{\textbf{For Potassium:}}}$
$$\lambda_0=\dfrac{hc}{E_0}=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(2.3\times 1.602\times 10^{-19})}=\color{red}{\bf 540}\;\rm nm$$
$\;\underline{\underline{\textbf{For Gold:}}}$
$$\lambda_0=\dfrac{hc}{E_0}=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(5.1\times 1.602\times 10^{-19})}=\color{red}{\bf 243}\;\rm nm$$
$$\color{blue}{\bf [c]}$$
$\;\underline{\underline{\textbf{For Potassium:}}}$
$$ v_{\rm max} =\sqrt{\dfrac{2\left(\dfrac{hc}{\lambda}-E_0\right)}{m}}\\
=\sqrt{\dfrac{2\left(\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(220\times 10^{-9})}-(2.3\times 1.602\times 10^{-19})\right)}{(9.11\times 10^{-31})}} $$
$$v_{\rm max}=\color{red}{\bf1.084\times 10^6}\;\rm m/s$$
$\;\underline{\underline{\textbf{For Gold:}}}$
$$ v_{\rm max} =\sqrt{\dfrac{2\left(\dfrac{hc}{\lambda}-E_0\right)}{m}}\\
=\sqrt{\dfrac{2\left(\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(220\times 10^{-9})}-(5.1\times 1.602\times 10^{-19})\right)}{(9.11\times 10^{-31})}} $$
$$v_{\rm max}=\color{red}{\bf 4.36\times 10^5}\;\rm m/s$$
$$\color{blue}{\bf [d]}$$
$\;\underline{\underline{\textbf{For Potassium:}}}$
$$V_{\rm stop}=\dfrac{\dfrac{hc}{\lambda}-E_0}{e} $$
$$ = \dfrac{ \dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(220\times 10^{-9})}-(2.3\times 1.602\times 10^{-19}) }{(1.6\times 10^{-19})}$$
$$V_{\rm stop}=\color{red}{\bf 3.34}\;\rm V$$
$\;\underline{\underline{\textbf{For Gold:}}}$
$$V_{\rm stop}=\dfrac{\dfrac{hc}{\lambda}-E_0}{e} $$
$$ = \dfrac{ \dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(220\times 10^{-9})}-(5.1\times 1.602\times 10^{-19}) }{(1.6\times 10^{-19})}$$
$$V_{\rm stop}=\color{red}{\bf 0.541}\;\rm V$$
