A student shines light of several known frequencies on two clean metal plates, A and B, in turn…

(a) Planck constant. Line A passes through $(5.5 \times 10^{14}\ \text{Hz}, 0)$ and $(9.0 \times 10^{14}\ \text{Hz}, 1.45\ \text{V})$. Its slope is $\dfrac{1.45\ \text{V}}{3.5 \times 10^{14}\ \text{Hz}} = 4.14 \times 10^{-15}\ \text{V s}$. This slope is $\dfrac{h}{e}$, so $h = 4.14 \times 10^{-15}\ \text{V s} \times 1.6 \times 10^{-19}\ \text{C} = 6.63 \times 10^{-34}\ \text{J s}$.

(b) and (c) Work functions. In electron volts, $\phi = (\text{slope in V s}) \times \nu_0$. Plate A: $4.14 \times 10^{-15}\ \text{V s} \times 5.5 \times 10^{14}\ \text{Hz} = 2.28\ \text{eV}$. Plate B: $4.14 \times 10^{-15}\ \text{V s} \times 7.5 \times 10^{14}\ \text{Hz} = 3.11\ \text{eV}$. The lines are parallel because they have the same slope, $\dfrac{h}{e}$, for every metal.

(d) Greatest kinetic energy. The frequency is $\nu = \dfrac{3.0 \times 10^{8}\ \text{m s}^{-1}}{420 \times 10^{-9}\ \text{m}} = 7.14 \times 10^{14}\ \text{Hz}$. This is above $5.5 \times 10^{14}\ \text{Hz}$ but below $7.5 \times 10^{14}\ \text{Hz}$: plate A emits, and plate B does not, because the photon energy $2.96\ \text{eV}$ is less than $3.11\ \text{eV}$. For plate A, $K_{\max} = 4.14 \times 10^{-15}\ \text{V s} \times (7.14 - 5.5) \times 10^{14}\ \text{Hz} = 0.68\ \text{eV}$.

(e) De Broglie wavelength. $K = 0.68\ \text{eV} \times 1.6 \times 10^{-19}\ \text{J eV}^{-1} = 1.09 \times 10^{-19}\ \text{J}$. The momentum is $p = \sqrt{2mK} = 4.45 \times 10^{-25}\ \text{kg m s}^{-1}$, so $\lambda = \dfrac{h}{p} = 1.49\ \text{nm}$. The fastest electron has the shortest wavelength.