Let's solve the problem step by step. --- **QUESTION RECAP:** A particle of mass \( m \) is confined to a two-dimensional infinite potential well defined by: - \( U(x, y) = 0 \) for \( 0 \leq x \leq 3L \) and \( 0 \leq y \leq 9L \) - \( U(x, y) = \infty \) otherwise **A. Using Schrödinger’s equation, derive a formula for the energy states of the particle.** ### **Step 1: Schrödinger Equation for 2D Infinite Well** The time-independent Schrödinger equation is: \[ -\frac{\hbar^2}{2m} \left( \frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} \right) = E \Psi \] ### **Step 2: Boundary Conditions** - \( \Psi(0, y) = \Psi(3L, y) = 0 \) - \( \Psi(x, 0) = \Psi(x, 9L) = 0 \) ### **Step 3: Separation of Variables** Assume \( \Psi(x, y) = X(x)Y(y) \): \[ -\frac{\hbar^2}{2m}\left( \frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} \right) = E \] Let \( E = E_x + E_y \). #### For \( x \): \[ \frac{d^2X}{dx^2} + k_x^2 X = 0 \] with boundary conditions \( X(0) = X(3L) = 0 \). Solution: \( X(x) = \sin\left(\frac{n_x \pi x}{3L}\right) \), where \( n_x = 1, 2, 3, \ldots \) So, \( k_x = \frac{n_x \pi}{3L} \) #### For \( y \): \[ \frac{d^2Y}{dy^2} + k_y^2 Y = 0 \] with boundary conditions \( Y(0) = Y(9L) = 0 \). Solution: \( Y(y) = \sin\left(\frac{n_y \pi y}{9L}\right) \), where \( n_y = 1, 2, 3, \ldots \) So, \( k_y = \frac{n_y \pi}{9L} \) ### **Step 4: Energy Formula** \[ E_{n_x, n_y} = \frac{\hbar^2}{2m} \left( k_x^2 + k_y^2 \right) = \frac{\hbar^2}{2m} \left[ \left( \frac{n_x \pi}{3L} \right)^2 + \left( \frac{n_y \pi}{9L} \right)^2 \right] = \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{n_x^2}{9} + \frac{n_y^2}{81} \right) \] --- **B. Identify the quantum numbers, energies, and degeneracy of the first two degenerate energy states that are integer multiples of \( E_0 = \frac{\hbar^2 \pi^2}{2m L^2} \).** ### **Step 1: Express Energy in terms of \( E_0 \)** \[ E_{n_x, n_y} = E_0 \left( \frac{n_x^2}{9} + \frac{n_y^2}{81} \right) \] We are to find the first two degenerate energy states (states with different quantum numbers but same energy), with energies that are integer multiples of \( E_0 \). ### **Step 2: Find Degenerate States** Let’s try some integer values for \( n_x \) and \( n_y \): #### **Try \( n_x = 3 \), \( n_y = 0 \) (but \( n_y \geq 1 \)), so try \( n_x = 3 \), \( n_y = 9 \):** \[ E_{3,9} = E_0 \left( \frac{9}{9} + \frac{81}{81} \right) = E_0 (1+1) = 2 E_0 \] #### **Try \( n_x = 6 \), \( n_y = 0 \) (not allowed), so try \( n_x = 6 \), \( n_y = 9 \):** \[ E_{6,9} = E_0 \left( \frac{36}{9} + \frac{81}{81} \right) = E_0 (4 + 1) = 5 E_0 \] #### **Try \( n_x = 9 \), \( n_y = 9 \):** \[ E_{9,9} = E_0 \left( \frac{81}{9} + \frac{81}{81} \right) = E_0 (9 + 1) = 10 E_0 \] #### **Try \( n_x = 6 \), \( n_y = 6\):** \[ E_{6,6} = E_0 \left( \frac{36}{9} + \frac{36}{81} \right) = E_0 (4 + 0.444) = 4.444 E_0 \] —not integer. #### **Try \( n_x = 0 \), \( n_y = 9 \): Not allowed; both quantum numbers must be ≥1.** #### **Try \( n_x = 9 \), \( n_y = 0 \): Not allowed.** #### **Try \( n_x = 6 \), \( n_y = 6 \): Already checked.** #### **Try \( n_x = 1 \), \( n_y = 9 \):** \[ E_{1,9} = E_0 \left( \frac{1}{9} + \frac{81}{81} \right) = E_0 \left( \frac{1}{9} + 1 \right) = E_0 \left( \frac{10}{9} \right) \] —not integer. #### **Try \( n_x = 3 \), \( n_y = 9 \):** Already got \( 2 E_0 \). #### **Try \( n_x = 6 \), \( n_y = 9 \):** Already got \( 5 E_0 \). #### **Try \( n_x = 9 \), \( n_y = 0 \), not allowed.** #### **Try \( n_x = 3 \), \( n_y = 18 \):** \[ E_{3,18} = E_0 \left( \frac{9}{9} + \frac{324}{81} \right) = E_0 (1 + 4) = 5 E_0 \] Degenerate with \( (6,9) \). #### **Try \( n_x = 6 \), \( n_y = 15 \):** \[ E_{6,15} = E_0 \left( \frac{36}{9} + \frac{225}{81} \right) = E_0 (4 + 2.778) = 6.778 E_0 \] —not integer. ### **Summarize:** #### **First two degenerate integer-multiple states:** 1. \( E = 2E_0 \): - \( (n_x, n_y) = (3, 9) \) and \( (9, 3) \) (But \( n_y \) must be ≤ 9, so only \( (3,9) \) valid in this well.) - DEGENERACY: 1 2. \( E = 5E_0 \): - \( (n_x, n_y) = (6,9) \) and \( (3,18) \) (but \( n_y \leq 9 \), so only \( (6,9) \) is valid) - DEGENERACY: 1 #### **Degeneracy:** In this well, due to the asymmetric boundaries, most states are non-degenerate (degeneracy 1), unless \( \frac{n_x^2}{9} = \frac{n_y^2}{81} \) and both quantum numbers are within allowed limits. --- ## **Final Answers** ### **A. Energy Formula** \[ \boxed{ E_{n_x, n_y} = \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{n_x^2}{9} + \frac{n_y^2}{81} \right) } \] where \( n_x = 1, 2, 3, \ldots \), \( n_y = 1, 2, 3, \ldots \) ### **B.** - **First degenerate integer-multiple state:** \( E = 2E_0 \), quantum numbers \( (n_x, n_y) = (3,9) \); degeneracy = 1 - **Second degenerate integer-multiple state:** \( E = 5E_0 \), quantum numbers \( (n_x, n_y) = (6,9) \); degeneracy = 1 If you have further questions about the degeneracy or want to check more states, let me know!