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

<p>We are given the work functions of Aluminium and Gold as $4.1 ~\mathrm{eV}$ and $5.1 ~\mathrm{eV}$, respectively.</p> <p>The stopping potential ($V_s$) is related to the frequency ($f$) of the incident light by the equation:</p> <p>$eV_s = h(f - f_0)$</p> <p>Where $e$ is the charge of an electron, $h$ is Planck&#39;s constant, and $f_0$ is the threshold frequency. The threshold frequency is related to the work function ($\phi$) by:</p> <p>$\phi = hf_0$</p> <p>So, we can write the equation for stopping potential as:</p> <p>$V_s = \frac{h}{e}(f - \frac{\phi}{h})$</p> <p>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:</p> <p>$$\frac{\text{Slope for Gold}}{\text{Slope for Aluminium}} = \frac{\frac{h}{e}}{\frac{h}{e}} = 1$$</p> <p>So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.</p>