Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2026 (4 Apr, Shift 1)

Answer: (a)

1. $eV_0 = h\nu - h\nu_0$; the axis intercept is the threshold frequency.
2. Thresholds: $X_1 = 1.0$, $X_2 = 1.5$, $X_3 = 2.0$ (in $10^{14}$ Hz), so $X_1$ has the smallest work function.
3. Same wavelength means same $h\nu$, so $KE_{max} = h\nu - \phi$ is largest for the smallest $\phi$.
4. So the answer is (a) $X_1$.