The work functions of Aluminium and Gold are 4.1 eV and 5.1 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

We need the ratio of slopes of stopping potential vs frequency plots for Gold and Aluminium.

State the photoelectric equation.

Einstein's photoelectric equation: $$eV_0 = h\nu - \phi$$

$$V_0 = \dfrac{h}{e}\nu - \dfrac{\phi}{e}$$

Identify the slope.

The slope of $$V_0$$ vs $$\nu$$ plot is $$\dfrac{h}{e}$$, which is a universal constant independent of the material.

Therefore, the ratio of slopes = $$\dfrac{h/e}{h/e} = 1$$.

This corresponds to Option 3. But the stored answer is Option 1 (1.24).

The slope $$h/e$$ is the same for all metals, so the ratio must be 1. The stored answer of 1.24 seems incorrect based on standard physics. However, if the question asks for something different (like ratio of threshold frequencies or cutoff wavelengths), the answer could differ.

The answer is $$\boxed{1}$$ (Option 3). The stored answer indicates Option 1.