Step 1 (Formula / Governing Law & Given Data): [1 Mark]
The photoelectric effect is governed by Einstein’s Photoelectric Equation, which relates the maximum kinetic energy ($K_{max}$) of emitted photoelectrons to the frequency of incident radiation ($\nu$) and the work function ($\Phi_0$) of the metal:
$$K_{max} = h\nu - \Phi_0$$Since the stopping potential ($V_s$) is defined as the potential required to stop the most energetic electrons, we have $K_{max} = eV_s$. Substituting this into the equation:
$$eV_s = h\nu - \Phi_0 \implies V_s = \left( \frac{h}{e} \right)\nu - \frac{\Phi_0}{e}$$This equation represents a straight line of the form $y = mx + c$, where the slope is $\frac{h}{e}$ and the y-intercept is $-\frac{\Phi_0}{e}$.
Step 2 (Step-by-Step Mathematical Working): [2 Marks]
- Slope Analysis: The slope of the $V_s$ versus $\nu$ graph is $\frac{h}{e}$. Since $h$ (Planck’s constant) and $e$ (charge of an electron) are universal constants, the slope for both metals A and B must be identical. The graph shows parallel lines, confirming this theoretical consistency.
- Work Function Deduction: The x-intercept of the graph represents the threshold frequency ($\nu_0$), where $V_s = 0$. At this point: $$0 = h\nu_0 - \Phi_0 \implies \Phi_0 = h\nu_0$$
- Comparison: By observing the x-intercepts on the graph, if $\nu_{0,B} > \nu_{0,A}$, then it follows that $\Phi_{0,B} > \Phi_{0,A}$. The metal with the larger x-intercept (further to the right) has a higher work function.
Step 3 (Final Concluded Answer): [1 Mark]
Based on the graphical analysis:
- Slope: The slope of the graph is equal to $\frac{h}{e}$, which is constant for all metals.
- Work Function: The metal with the higher threshold frequency ($\nu_0$) has a higher work function ($\Phi_0$). Therefore, if the intercept for B is greater than A, then Metal B has a higher work function than Metal A.
💡 EL10 Concept Insight:
Think of the work function as the "entry fee" a light particle must pay to kick an electron out of the metal. The graph shows that Metal B needs a higher frequency (more energy) just to pay that fee, meaning it holds onto its electrons more tightly than Metal A.