In a photoelectric experiment using a particular metal, the stopping potential is $V_{01} = 1.5\,\text{V}$ when the incident frequency is $\nu_1 = 8\times 10^{14}\,\text{Hz}$, and $V_{02} = 3.2\,\text{V}$ when $\nu_2 = 1.2\times 10^{15}\,\text{Hz}$. Using only this data, the value of Planck's constant $h$ is closest to (Take $e = 1.6\times 10^{-19}\,\text{C}$.)

1. Einstein's photoelectric equation: $eV_0 = h\nu - \phi_0$. 2. Write the equation for both frequencies: $eV_{01} = h\nu_1 - \phi_0$ and $eV_{02} = h\nu_2 - \phi_0$. 3. SUBTRACT to eliminate $\phi_0$: $e(V_{02} - V_{01}) = h(\nu_2 - \nu_1)$. 4. Solve for $h$: $h = \dfrac{e(V_{02} - V_{01})}{\nu_2 - \nu_1}$. 5. Plug in: $V_{02} - V_{01} = 1.7\,\text{V}$; $\nu_2 - \nu_1 = 4\times 10^{14}\,\text{Hz}$. 6. $h = \dfrac{(1.6\times 10^{-19})(1.7)}{4\times 10^{14}} = \dfrac{2.72\times 10^{-19}}{4\times 10^{14}} = 6.8\times 10^{-34}\,\text{J\,s}$. 7. This agrees with the accepted value of Planck's constant. Distractors come from arithmetic slips: A uses $V_{02}+V_{01}$ in the wrong sign; B and C are intermediate miscalculations. _Source: NCERT Class 12 Physics Part 2, Ch 11, §11.6 Fig. 11.5 (V₀ vs ν line — Millikan's method for finding $h$), p. 7._