An electron is confined to a metal cube of \(l=0.8 \mathrm{cm}\) (a) on each side. Determine the density of states at \(E=0.80 \mathrm{eV} ;\) (b) \(E=2.2 \mathrm{eV} ;\) and \((\mathrm{c}) E=5.0 \mathrm{eV}\)

The volume of a cube with side length l can be found using the formula: \[ V = l^3 \] Given that the side length l is 0.8 cm, convert it to meters. \[ l = 0.8 \times 10^{-2} \mathrm{m} \] Now, calculate the volume. \[ V = (0.8 \times 10^{-2})^3 = 5.12 \times 10^{-7} \mathrm{m^3} \]

Now we will use the volume and the energy levels to find the density of states using the approximate formula: \[ D(E) \approx \frac{V}{(2\pi)^2}\left(\frac{2m}{\hbar^2}\right)^{\frac{3}{2}} E^{\frac{1}{2}} \] where \( m = 9.11 \times 10^{-31} \mathrm{kg} \) is the mass of the electron and \( \hbar = 1.055 \times 10^{-34} \mathrm{J}\cdot\mathrm{s} \) is the reduced Planck's constant. For each energy level (0.80 eV, 2.2 eV, and 5.0 eV), convert the energy to joules and use it to find the density of states: (a) E = 0.80 eV = \( 0.80 \times 1.602 \times 10^{-19} \mathrm{J} \) \[ D(0.80\mathrm{eV}) \approx \frac{5.12 \times 10^{-7}}{(2\pi)^2}\left(\frac{2(9.11 \times 10^{-31})}{(1.055 \times 10^{-34})^2}\right)^{\frac{3}{2}} (0.80 \times 1.602 \times 10^{-19})^{\frac{1}{2}} \] (b) E = 2.2 eV = \( 2.2 \times 1.602 \times 10^{-19} \mathrm{J} \) \[ D(2.2\mathrm{eV}) \approx \frac{5.12 \times 10^{-7}}{(2\pi)^2}\left(\frac{2(9.11 \times 10^{-31})}{(1.055 \times 10^{-34})^2}\right)^{\frac{3}{2}} (2.2 \times 1.602 \times 10^{-19})^{\frac{1}{2}} \] (c) E = 5.0 eV = \( 5.0 \times 1.602 \times 10^{-19} \mathrm{J} \) \[ D(5.0\mathrm{eV}) \approx \frac{5.12 \times 10^{-7}}{(2\pi)^2}\left(\frac{2(9.11 \times 10^{-31})}{(1.055 \times 10^{-34})^2}\right)^{\frac{3}{2}} (5.0 \times 1.602 \times 10^{-19})^{\frac{1}{2}} \] Calculate the density of states for each energy level and write down the results.