We are given the work functions of Aluminium and Gold as $$4.1 ~\mathrm{eV}$$ and $$5.1 ~\mathrm{eV}$$, respectively.

The stopping potential ($$V_s$$) is related to the frequency ($$f$$) of the incident light by the equation:

$$eV_s = h(f - f_0)$$

Where $$e$$ is the charge of an electron, $$h$$ is Planck's constant, and $$f_0$$ is the threshold frequency. The threshold frequency is related to the work function ($$\phi$$) by:

$$\phi = hf_0$$

So, we can write the equation for stopping potential as:

$$V_s = \frac{h}{e}(f - \frac{\phi}{h})$$

This equation represents a straight line with slope $$\frac{h}{e}$$. Therefore, the slope of the stopping potential versus frequency plot is the same for both Aluminium and Gold. The ratio of the slopes is:

$$\frac{\text{Slope for Gold}}{\text{Slope for Aluminium}} = \frac{\frac{h}{e}}{\frac{h}{e}} = 1$$

So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.