Let's solve for the energy levels of an electron in a **finite square well** of: - Width \( L = 0.5\,\text{nm} = 0.5 \times 10^{-9}\,\text{m} \) - Well depth \( V_0 = 5.0\,\text{eV} \) - Electron mass \( m = 9.11 \times 10^{-31}\,\text{kg} \) - \( \hbar = 1.055 \times 10^{-34}\,\text{J}\cdot\text{s} \) - \( 1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J} \) We are to **iteratively solve** for the allowed energies. --- ## 1. The Transcendental Equations For a symmetric well (\(-L/2 < x < L/2\)), the bound state energies are given by: Let \( E \) be the bound state energy (\( 0 < E < V_0 \)), and define: \[ k = \sqrt{\frac{2mE}{\hbar^2}} \] \[ \kappa = \sqrt{\frac{2m(V_0-E)}{\hbar^2}} \] The transcendental equations for even and odd states are: - **Even states** (ground state, etc.): \[ k \tan\left(\frac{kL}{2}\right) = \kappa \] - **Odd states** (first excited, etc.): \[ -k \cot\left(\frac{kL}{2}\right) = \kappa \] --- ## 2. Dimensionless Form Define: \[ z = \frac{kL}{2}, \qquad z_0 = \frac{L}{2} \sqrt{\frac{2mV_0}{\hbar^2}} \] Then, \[ \kappa = \sqrt{\frac{2m(V_0-E)}{\hbar^2}} = \sqrt{\frac{2mV_0}{\hbar^2} - \frac{2mE}{\hbar^2}} = \frac{2}{L} \sqrt{z_0^2 - z^2} \] So: - **Even**: \(\tan z = \frac{\sqrt{z_0^2 - z^2}}{z} \) - **Odd**: \(-\cot z = \frac{\sqrt{z_0^2 - z^2}}{z} \) --- ## 3. Calculate \(z_0\): First, calculate \(z_0\): \[ z_0 = \frac{L}{2} \sqrt{\frac{2mV_0}{\hbar^2}} \] Plug in values: - \(L/2 = 0.25 \times 10^{-9}\,\text{m}\) - \(2m = 2 \times 9.11 \times 10^{-31} = 1.822 \times 10^{-30}\) - \(V_0 = 5.0\,\text{eV} = 8.01 \times 10^{-19}\,\text{J}\) - \(\hbar^2 = (1.055 \times 10^{-34})^2 = 1.113 \times 10^{-68}\) Calculate the inside of the square root: \[ \frac{2mV_0}{\hbar^2} = \frac{1.822 \times 10^{-30} \times 8.01 \times 10^{-19}}{1.113 \times 10^{-68}} = \frac{1.459 \times 10^{-48}}{1.113 \times 10^{-68}} = 1.311 \times 10^{20} \] Therefore, \[ z_0 = 0.25 \times 10^{-9} \times \sqrt{1.311 \times 10^{20}} = 0.25 \times 10^{-9} \times 1.145 \times 10^{10} = 2.86 \] --- ## 4. **Find the allowed \( z \) values (iterative approach)** The allowed \(z\) values are solutions to: - Even: \( \tan z = \frac{\sqrt{z_0^2 - z^2}}{z} \) - Odd: \( -\cot z = \frac{\sqrt{z_0^2 - z^2}}{z} \) But \(z\) must be less than \(z_0 = 2.86\). ### a) **Ground state (even solution):** We look for a solution between \(z = 0\) and \(z = \pi/2 \approx 1.57\). Set up the function: \[ f(z) = \tan z - \frac{\sqrt{z_0^2 - z^2}}{z} \] We seek \(f(z) = 0\). Let's try some values numerically: #### Try \( z = 1.0 \): \[ \frac{\sqrt{2.86^2 - 1.0^2}}{1.0} = \frac{\sqrt{8.18 - 1.0}}{1.0} = \frac{\sqrt{7.18}}{1.0} \approx 2.68 \] \(\tan(1.0) = 1.557\) So \(f(1.0) = 1.557 - 2.68 = -1.123\) #### Try \( z = 1.3 \): \[ \sqrt{2.86^2 - 1.69} = \sqrt{8.18 - 1.69} = \sqrt{6.49} = 2.55 \] \(2.55 / 1.3 = 1.96\) \(\tan(1.3) = 3.602\) So \(f(1.3) = 3.602 - 1.96 = 1.64\) So between \(z = 1.0\) (f = -1.12) and \(1.3\) (f = +1.64), the function crosses zero. #### Try \( z = 1.15 \): \[ \sqrt{8.18 - 1.3225} = \sqrt{6.8575} = 2.62 \] \(2.62 / 1.15 = 2.28\) \(\tan(1.15) = 2.1445\) \(f(1.15) = 2.1445 - 2.28 = -0.135\) #### Try \( z = 1.2 \): \[ \sqrt{8.18 - 1.44} = \sqrt{6.74} = 2.60 \] \(2.60 / 1.2 = 2.17\) \(\tan(1.2) = 2.5721\) \(f(1.2) = 2.5721 - 2.17 = 0.40\) So at \(z = 1.15\), \(f = -0.135\); at \(z = 1.2\), \(f = 0.40\). Interpolate: \[ z_1 \approx 1.15 + \frac{0.135}{0.135+0.40} \times (1.2 - 1.15) = 1.15 + \frac{0.135}{0.535} \times 0.05 \approx 1.15 + 0.013 = 1.163 \] Let’s check at \(z = 1.17\): \[ \sqrt{8.18 - 1.3689} = \sqrt{6.8111} = 2.61 \] \(2.61/1.17 = 2.23\) \(\tan(1.17) = 2.302\) \(f(1.17) = 2.302 - 2.23 = 0.072\) Now at \(z=1.15\): \(f=-0.135\), \(z=1.17\): \(f=0.072\). So zero crossing is at \[ z = 1.15 + \frac{0.135}{0.135+0.072} \times (1.17-1.15) \approx 1.15 + \frac{0.135}{0.207} \times 0.02 \approx 1.15 + 0.013 = 1.163 \] **So, the ground state solution is at \( z_1 \approx 1.16 \) (to three sig. figs).** --- ### b) **First excited state (odd solution):** Odd solution: \( -\cot z = \frac{\sqrt{z_0^2 - z^2}}{z} \) So, \( f(z) = -\cot z - \frac{\sqrt{z_0^2 - z^2}}{z} \) Odd solutions are between \(z = \pi/2\) and \(z_0\): - \(\pi/2 \approx 1.57\) - \(z_0 = 2.86\) Let’s try \(z = 2.0\): \[ \sqrt{8.18 - 4.0} = \sqrt{4.18} = 2.04 \] \(2.04 / 2.0 = 1.02\) \(-\cot(2.0) = -1/\tan(2.0) = -1/(-2.185) = 0.457\) \(f(2.0) = 0.457 - 1.02 = -0.563\) Try \(z = 2.4\): \[ \sqrt{8.18 - 5.76} = \sqrt{2.42} = 1.56 \] \(1.56 / 2.4 = 0.65\) \(\tan(2.4) = -0.883\), \(-\cot(2.4) = -1/(-0.883) = 1.13\) \(f(2.4) = 1.13 - 0.65 = 0.48\) So, at \(z=2.0\), \(f=-0.563\); at \(z=2.4\), \(f=0.48\). Try \(z=2.2\): \[ \sqrt{8.18 - 4.84} = \sqrt{3.34} = 1.83 \] \(1.83/2.2=0.83\) \(\tan(2.2) = -1.370\), \(-\cot(2.2) = -1/(-1.37)=0.73\) \(f(2.2) = 0.73 - 0.83 = -0.10\) Try \(z=2.3\): \[ \sqrt{8.18 - 5.29} = \sqrt{2.89} = 1.70 \] \(1.70/2.3 = 0.739\) \(\tan(2.3) = -1.342\), \(-\cot(2.3) = -1/(-1.342) = 0.745\) \(f(2.3) = 0.745 - 0.739 = 0.006 \] Very close to zero! Let's check at \(z=2.28\): \[ \sqrt{8.18 - 5.1984} = \sqrt{2.9816} = 1.727 \] \(1.727/2.28=0.758\) \(\tan(2.28) = -1.381\), \(-\cot(2.28) = -1/(-1.381) = 0.724\) \(f(2.28) = 0.724 - 0.758 = -0.034\) So, at \(z=2.28\), \(f=-0.034\); at \(z=2.3\), \(f=0.006\). So crossing is at \[ z_2 = 2.28 + \frac{0.034}{0.034+0.006} \times (2.3-2.28) = 2.28 + \frac{0.034}{0.04} \times 0.02 = 2.28 + 0.017 = 2.297 \] So, to three significant figures, \(z_2 = 2.30\). --- ## 5. **Convert \(z\) to Energy** Recall: \[ z = \frac{kL}{2} = \frac{L}{2} \sqrt{\frac{2mE}{\hbar^2}} \] So, \[ E = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m} \left(\frac{2z}{L}\right)^2 = \frac{2\hbar^2 z^2}{mL^2} \] Plug in numbers: - \(\hbar^2 = 1.113 \times 10^{-68}\) - \(m = 9.11 \times

Let's carefully compute the energies of the ground and first excited states of an electron in a finite potential well of width 0.5 nm and height 5.0 eV using an iterative approach. --- ### Step 1: Convert known quantities to SI units - Well width: \(L = 0.5\,\text{nm} = 0.5 \times 10^{-9}\,\text{m}\) - Well depth: \(V_0 = 5.0\,\text{eV} = 5.0 \times 1.602 \times 10^{-19}\,\text{J} = 8.01 \times 10^{-19}\,\text{J}\) - Electron mass: \(m = 9.11 \times 10^{-31}\,\text{kg}\) - Reduced Planck's constant: \(\hbar = 1.055 \times 10^{-34}\,\text{Js}\) --- ### Step 2: Calculate the parameter \(z_0\) The transcendental equations are best handled in dimensionless form. Define: \[ z_0 = \frac{L}{2} \sqrt{\frac{2mV_0}{\hbar^2}} \] Compute: \[ \frac{2mV_0}{\hbar^2} = \frac{2 \times 9.11 \times 10^{-31} \times 8.01 \times 10^{-19}}{(1.055 \times 10^{-34})^2} \] Calculate numerator: \[ 2 \times 9.11 \times 10^{-31} \times 8.01 \times 10^{-19} \approx 1.459 \times 10^{-48} \] Calculate denominator: \[ (1.055 \times 10^{-34})^2 \approx 1.113 \times 10^{-68} \] Thus: \[ \frac{2mV_0}{\hbar^2} \approx \frac{1.459 \times 10^{-48}}{1.113 \times 10^{-68}} \approx 1.311 \times 10^{20} \] Now: \[ z_0 = \frac{0.5 \times 10^{-9}}{2} \times \sqrt{1.311 \times 10^{20}} = 0.25 \times 10^{-9} \times 1.145 \times 10^{10} \approx 2.86 \] --- ### Step 3: Find the allowed \(z\) values The solutions satisfy: - **Even states:** \[ \tan z = \frac{\sqrt{z_0^2 - z^2}}{z} \] - **Odd states:** \[ -\cot z = \frac{\sqrt{z_0^2 - z^2}}{z} \] with \(z < z_0 \approx 2.86\). --- ### Step 4: Find the ground state energy (even solution) Estimate \(z\) for the ground state: - The first solution occurs between \(z=0\) and \(\pi/2 \approx 1.57\). Let's iterate: **At \(z=0.5\):** \[ \sqrt{2.86^2 - 0.5^2} = \sqrt{8.18 - 0.25} = \sqrt{7.93} \approx 2.82 \] \[ \frac{\sqrt{z_0^2 - z^2}}{z} \approx \frac{2.82}{0.5} = 5.64 \] \[ \tan(0.5) \approx 0.546 \] Since \(0.546 < 5.64\), the RHS is larger; increase \(z\). **At \(z=1.0\):** \[ \sqrt{8.18 - 1} = \sqrt{7.18} \approx 2.68 \] \[ 2.68/1.0 = 2.68 \] \[ \tan(1.0) \approx 1.557 \] Since \(1.557 < 2.68\), increase \(z\). **At \(z=1.2\):** \[ \sqrt{8.18 - 1.44} = \sqrt{6.74} \approx 2.6 \] \[ 2.6/1.2 \approx 2.17 \] \[ \tan(1.2) \approx 2.572 \] Since \(2.572 > 2.17\), the LHS exceeds RHS; decrease \(z\). Between 1.0 and 1.2, at \(z=1.1\): \[ \sqrt{8.18 - 1.21} = \sqrt{6.97} \approx 2.64 \] \[ 2.64/1.1 \approx 2.4 \] \(\tan(1.1) \approx 1.964\) Since \(1.964 < 2.4\), increase \(z\). At \(z=1.15\): \[ \sqrt{8.18 - 1.3225} = \sqrt{6.8575} \approx 2.62 \] \[ 2.62/1.15 \approx 2.28 \] \(\tan(1.15) \approx 2.1445\) Since \(2.1445 < 2.28\), increase \(z\). At \(z=1.17\): \[ \sqrt{8.18 - 1.3689} = \sqrt{6.8111} \approx 2.61 \] \[ 2.61/1.17 \approx 2.23 \] \(\tan(1.17) \approx 2.572\) Now, \(\tan z \approx 2.572\) and RHS \(\approx 2.23\). The LHS exceeds RHS, so solution near \(z \approx 1.16\). **Interpolating:** Between 1.15 and 1.17, the solution is approximately **\(z \approx 1.16\)**. --- ### Step 5: Calculate the ground state energy \[ E_1 = \frac{\hbar^2 k^2}{2m} \] where \[ k = \frac{2z}{L} \] So: \[ E_1 = \frac{\hbar^2}{2m} \left(\frac{2z}{L}\right)^2 = \frac{4 \hbar^2 z^2}{2 m L^2} = \frac{2 \hbar^2 z^2}{m L^2} \] Plug in: \[ \frac{2 \times (1.055 \times 10^{-34})^2 \times (1.16)^2}{9.11 \times 10^{-31} \times (0.5 \times 10^{-9})^2} \] Calculate numerator: \[ 2 \times 1.113 \times 10^{-68} \times 1.3456 \approx 2.996 \times 10^{-68} \] Calculate denominator: \[ 9.11 \times 10^{-31} \times 0.25 \times 10^{-18} = 2.278 \times 10^{-49} \] Thus: \[ E_1 \approx \frac{2.996 \times 10^{-68}}{2.278 \times 10^{-49}} \approx 1.315 \times 10^{-19}\,\text{J} \] Convert to eV: \[ E_1 = \frac{1.315 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 0.82\,\text{eV} \] ### **Ground state energy: \(\boxed{0.82\,\text{eV}}\)** --- ### **Step 6: Find the first excited state (odd solution)** Using similar procedure: Solutions occur between \(\pi/2 \approx 1.57\) and \(z_0=2.86\). Test at \(z=2.3\): \[ \sqrt{8.18 - 2.3^2} = \sqrt{8.18 - 5.29} = \sqrt{2.89} \approx 1.70 \] \[ \frac{1.70}{2.3} \approx 0.739 \] \[ -\cot(2.3) \approx -(-1/ \tan 2.3) \approx -(-1/(-1.370)) \approx -0.730 \] Calculate \(f(z)\): \[ f(2.3) = -\cot z - \frac{\sqrt{z_0^2 - z^2}}{z} \approx 0.730 - 0.739 = -0.009 \] At \(z = 2.31\): \[ \sqrt{8.18 - 2.34^2} = \sqrt{8.18 - 5.48} = \sqrt{2.70} \approx 1.64 \] \[ 1.64/2.31 \approx 0.71 \] \(\tan 2.31 \approx -1.305\), so: \[ -\cot 2.31 = -(-1/ -1.305) \approx -0.766 \] \(f(2.31) = 0.766 - 0.71 = 0.056\) Zero crossing between \(z=2.3\) and 2.31. Interpolating: \[ z \approx 2.30 + \frac{0.009}{0.009 + 0.056} \times 0.01 \approx 2.30 + 0.0015 = 2.3015 \] Thus, \(z \approx 2.30\) (to three sig. figs). --- ### Step 7: Calculate the excited state energy \[ E_2 = \frac{2 \hbar^2 z^2}{m L^2} \] Using \(z=2.30\): \[ E_2 \approx \frac{2 \times 1.113 \times 10^{-68} \times (2.30)^2}{9.11 \times 10^{-31} \times (0.5 \times 10^{-9})^2} \] Numerator: \[ 2 \times 1.113 \times 10^{-68} \times 5.29 \approx 11.77 \times 10^{-68} \] Denominator: \[ 9.11 \times 10^{-31} \times 0.25 \times 10^{-18} = 2.278 \times 10^{-49} \] Calculate: \[ E_2 \approx \frac{11.77 \times 10^{-68}}{2.278 \times 10^{-49}} \approx 5.17 \times 10^{-19}\,\text{J} \] Convert to eV: \[ E_2 \approx \frac{5.17 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 3.23\,\text{eV} \] --- ### **Final Results:** | State | Energy (eV) | |---------|--------------| | Ground state | **0.82 eV** | | First excited state | **3.23 eV** | --- ### **Answer:** **The approximate energies are:** \[ \boxed{ \begin{aligned} E_1 &\approx \boxed{0.82\,\text{eV}} \quad \text{(ground state)} \\ E_2 &\approx \boxed{3.23\,\text{eV}} \quad \text{(first excited state)} \\ \end{aligned} } \] These are consistent with the iterative solutions and typical finite well behavior.