The work functions of Aluminium and Gold are 4 . 1   eV and 5 . 1   eV respectively. The ratio of…

The data given is 

${\varphi }_{Al}=4.1 \mathrm{eV}$

${\varphi }_{Au}=5.1 \mathrm{eV}$

The energy is given by 

$$\begin{gathered} h\nu =\varphi +KE \\ \Rightarrow h\nu =\varphi +e{V}_0 \end{gathered}$$

where $V_0$ is the stopping potential and $K=e{V}_0$

Thus, the above equation can be written as

$e{V}_0=h\nu -\varphi$

$\Rightarrow V_0=\frac{h\nu }{e}-\frac{\varphi }{e}$

The equation is of the form of $y=mx+c$. The slope is 

$m=\frac{h}{e}$.

Since the Planck constant $h$ and charge of an electron is a constant $e$, the slope does not depend on the work function. Hence the ratio of the slope of the plot of stopping potential and frequency for both the given metals is $1$.