An electron is confined to a finite well of width 0.5 nm and height 5.0 eV. Use an iterative approach to find the energy of the first ground state and first excited state to three significant figures.

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An electron is confined to a finite well of width 0.5 nm and height 5.0 eV. Use an iterative approach to find the energy of the first ground state and first excited state to three significant figures.                                                 give answer in 3 step with explanation at the end of each step and give final answer at the end

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Let's solve for the ground state and first excited state energy of an electron in a **finite potential well** of width $ L = 0.5\, \text{nm} $ and height $ V_0 = 5.0\, \text{eV} $, **using an iterative approach**. --- ## **STEP 1: Write Down the Transcendental Equations** For a finite well, the energies are found by solving the following equations for even (ground state) and odd (first excited state) solutions: Let $ E $ be the energy of the state, $ V_0 $ the well depth, and $ L $ the well width. Define: - $ k = \sqrt{\frac{2mE}{\hbar^2}} $ - $ \kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} $ The quantization conditions: - **Even states:** $ k\, L = 2 \tan^{-1} \left( \frac{\kappa}{k} \right) $ - **Odd states:** $ k\, L = 2 \tan^{-1} \left( \frac{k}{\kappa} \right) $ Or, more commonly written: - **Even:** $ \tan\left(\frac{k L}{2}\right) = \frac{\kappa}{k} $ - **Odd:** $ \tan\left(\frac{k L}{2}\right) = -\frac{k}{\kappa} $ **Explanation:** These equations relate the energy $ E $ (via $ k $ and $ \kappa $) to the width and depth of the well. We need to solve these equations for $ E $. --- ## **STEP 2: Substitute Physical Constants and Simplify Expressions** We need to express $ k $ and $ \kappa $ in terms of $ E $ (in eV): - Electron mass $ m = 9.11 \times 10^{-31}\,\text{kg} $ - Planck's reduced constant $ \hbar = 1.055 \times 10^{-34}\,\text{J}\cdot\text{s} $ - $ 1\,\text{eV} = 1.602 \times 10^{-19}\,\text{J} $ - $ L = 0.5\,\text{nm} = 0.5 \times 10^{-9}\,\text{m} $ Convert all energies to **joules** for calculation. Let’s define a constant: $$ \alpha = \frac{2m}{\hbar^2} = \frac{2 \times 9.11 \times 10^{-31}}{(1.055 \times 10^{-34})^2} \approx 1.637 \times 10^{20}\,\text{J}^{-1}\,\text{m}^{-2} $$ But let's work in **eV and nm** for simplicity. In these units: $$ k = \frac{\sqrt{2mE}}{\hbar} $$ $$ \hbar c = 197.3\, \text{eV}\cdot \text{nm} $$ $$ k = \frac{\sqrt{2m_e E}}{\hbar} = 5.12 \times 10^9 \sqrt{E(\text{eV})}\, \text{m}^{-1} $$ But more simply, in nm$^{-1}$: $$ k = \frac{\sqrt{2 m_e E}}{\hbar} $$ with $$ k (\text{nm}^{-1}) = 5.12 \sqrt{E (\text{eV})} $$ and $$ \kappa (\text{nm}^{-1}) = 5.12 \sqrt{V_0 - E} $$ **Explanation:** We now have all variables in convenient units. The transcendental equations can now be iteratively solved for $ E $ using these expressions for $ k $ and $ \kappa $. --- ## **STEP 3: Iteratively Solve for $ E $ (Numerically)** **Even (ground state):** $$ \tan\left(\frac{k L}{2}\right) = \frac{\kappa}{k} $$ with $$ k = 5.12 \sqrt{E} $$ $$ \kappa = 5.12 \sqrt{V_0 - E} $$ $$ L = 0.5\, \text{nm} $$ Let $ \theta = \frac{k L}{2} = 1.28 \sqrt{E} $ So, $$ \tan(1.28 \sqrt{E}) = \sqrt{\frac{V_0 - E}{E}} $$ **Let's solve this equation for $ V_0 = 5.0\,\text{eV} $.** ### **Ground State ($ n=1 $)** Let’s try values of $ E $ between 0 and 5 eV. #### Try $ E = 1.0\,\text{eV} $: $$ \tan(1.28 \times 1.0) = \tan(1.28) \approx 3.06 $$ $$ \sqrt{\frac{5 - 1}{1}} = \sqrt{4} = 2 $$ LHS (3.06) > RHS (2), so try a higher $ E $. #### Try $ E = 1.5\,\text{eV} $: $$ \tan(1.28 \times \sqrt{1.5}) = \tan(1.28 \times 1.225) = \tan(1.568) \approx 51.2 $$ But that's greater than the earlier value; but let's check calculation: Wait, the equation is $\tan(1.28\sqrt{E})$, so for $E = 1.5$: $$ \sqrt{1.5} = 1.225 $$ $$ 1.28 \times 1.225 = 1.568 $$ $$ \tan(1.568) \approx 51.2 $$ $$ \sqrt{\frac{5 - 1.5}{1.5}} = \sqrt{\frac{3.5}{1.5}} = \sqrt{2.333} = 1.528 $$ Again, LHS much greater than RHS. Try higher $ E $. But wait, as $ E $ increases, $\tan(1.28\sqrt{E})$ increases rapidly, so we need to try lower $ E $. Actually, at very low $ E $, $\tan(1.28\sqrt{E})$ is small, so let's try $ E = 0.5 $: $$ \sqrt{0.5} = 0.707 $$ $$ 1.28 \times 0.707 = 0.905 $$ $$ \tan(0.905) = 1.27 $$ $$ \sqrt{\frac{5 - 0.5}{0.5}} = \sqrt{9} = 3 $$ Now, LHS (1.27) < RHS (3). So, as $ E $ increases from 0.5 to 1.0 eV, LHS increases from 1.27 to 3.06, while RHS decreases from 3 to 2. So, the solution is between $ E = 0.5 $ and $ E = 1.0 $. Try $ E = 0.75 $: $$ \sqrt{0.75} = 0.866 $$ $$ 1.28 \times 0.866 = 1.108 $$ $$ \tan(1.108) = 1.964 $$ $$ \sqrt{\frac{5 - 0.75}{0.75}} = \sqrt{\frac{4.25}{0.75}} = \sqrt{5.667} = 2.382 $$ LHS (1.964), RHS (2.382) → LHS < RHS. Try $ E = 0.9 $: $$ \sqrt{0.9} = 0.949 $$ $$ 1.28 \times 0.949 = 1.215 $$ $$ \tan(1.215) = 2.696 $$ $$ \sqrt{\frac{5 - 0.9}{0.9}} = \sqrt{\frac{4.1}{0.9}} = \sqrt{4.556} = 2.136 $$ LHS (2.696) > RHS (2.136). So, the solution is between $ E = 0.75 $ and $ E = 0.9 $. Try $ E = 0.8 $: $$ \sqrt{0.8} = 0.894 $$ $$ 1.28 \times 0.894 = 1.144 $$ $$ \tan(1.144) = 2.070 $$ $$ \sqrt{\frac{5 - 0.8}{0.8}} = \sqrt{\frac{4.2}{0.8}} = \sqrt{5.25} = 2.291 $$ LHS (2.070), RHS (2.291). Try $ E = 0.85 $: $$ \sqrt{0.85} = 0.922 $$ $$ 1.28 \times 0.922 = 1.181 $$ $$ \tan(1.181) = 2.336 $$ \[ \sqrt{\frac{5 - 0.85}{0.85}} = \sqrt{\frac{4.15}{0.85}} = \sqrt{4.882}

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An electron is confined to a finite well of width 0.5 nm and height 5.0 eV. Use an iterative approach to find the energy of the first ground state and first excited state to three significant figures. give answer in 3 step with explanation at the end of each step and give final answer at the end

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