A student measures the stopping potential of a metal N…

Step 1. The slope. $\dfrac{1.30\ \text{V} - 0.30\ \text{V}}{(10.4 - 8.0) \times 10^{14}\ \text{Hz}} = 4.17 \times 10^{-15}\ \text{V s}$.

Step 2. Planck's constant. $h = e \times \text{slope} = 1.6 \times 10^{-19}\ \text{C} \times 4.17 \times 10^{-15}\ \text{V s} = 6.67 \times 10^{-34}\ \text{J s}$.

Step 3. The threshold. At the first point, $0.30\ \text{V} = 4.17 \times 10^{-15}\ \text{V s} \times (8.0 \times 10^{14}\ \text{Hz} - \nu_0)$. So $8.0 \times 10^{14}\ \text{Hz} - \nu_0 = 0.72 \times 10^{14}\ \text{Hz}$, and $\nu_0 = 7.28 \times 10^{14}$ Hz.

Step 4. The work function. $\phi = 4.17 \times 10^{-15}\ \text{V s} \times 7.28 \times 10^{14}\ \text{Hz} = 3.03\ \text{eV}$.