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

The data given is

${ϕ}_{Al} = 4 . 1eV$

${ϕ}_{Au} = 5 . 1eV$

The energy is given by

$hν = ϕ + KE ⇒ hν = ϕ + e{V}_{0}$

where ${V}_{0}$ is the stopping potential and $K = e{V}_{0}$

Thus, the above equation can be written as

$e{V}_{0} = hν - ϕ$

$⇒ {V}_{0} = \dfrac{hν}{e} - \dfrac{ϕ}{e}$

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

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

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 $1$.