QJEEM23Phcb2d: JEE Main 2023 Physics - Class 12 JEEM23Phcb2d Solution: The work functions of Aluminium and Gold are 4.1   eV and…

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:

$$e{V}_s=h(f-f_0)$$

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 ($$\varphi$$) by:

$$\varphi =h{f}_0$$

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

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

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.