Sol. 1. From Einstein's photoelectric equation: $$ K_{\max }=h v-\phi $$ where $K_{\max }$ is the maximum kinetic energy, $h$ is Planck's constant, $v$ is the frequency of incident radiation, and $\phi$ is the work function. 2. The work function $\phi$ is related to the threshold frequency $v_0$ by: $$ \phi=h v_0 $$ 3. Substituting $\phi$ into the kinetic energy equation: $$ K_{\max }=h v-h v_0 $$ 4. For a constant incident wavelength $\lambda$ (and thus constant frequency $\nu=c / \lambda$ ): $$ K_{\max } \propto-v_0 $$ This implies that the metal with the lowest threshold frequency $v_0$ will yield electrons with the highest maximum kinetic energy. 5. From the provided $V_0$ vs $v$ graph, the x-intercepts represent the threshold frequencies: - For $X_1: v_{0,1}=1.0 \times 10^{14} \mathrm{~Hz}$ - For $X_2: v_{0,2}=1.5 \times 10^{14} \mathrm{~Hz}$ - For $X_3: v_{0,3}=2.0 \times 10^{14} \mathrm{~Hz}$ 6. Comparing the values: $$ v_{0,1}