Example 37.1 calculates the energy of the wave function with the lowest quantum number for an electron confined to a box of width \(2.00 \AA\) in the one-dimensional case. However, atoms are three-dimensional entities with a typical diameter of \(1.00 \AA=10^{-10} \mathrm{~m} .\) It would seem then that the next, better approximation would be that of an electron trapped in a three-dimensional infinite potential well (a potential cube with sides of \(1.00 \mathrm{~A}\) ). a) Derive an expression for the electron wave function and the corresponding energies for a particle in a three dimensional rectangular infinite potential well. b) Calculate the lowest energy allowed for the electron in this case.

Consider an electron trapped in an infinite potential well of dimensions \(L_x = L_y = L_z = 1.00 \mathrm{~A} = 10^{-10}\mathrm{~m}\). For the particle to be confined, the potential energy must be: - V(x, y, z) = 0 for 0 < x < \(L_x\), 0 < y < \(L_y\), 0 < z < \(L_z\) - V(x, y, z) = ∞ for any other x, y, or z values. The Schrödinger equation that governs the particle's wave function is given by: \(-\frac{\hbar^2}{2m}(\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2})+V(x,y,z)\Psi = E\Psi\) For our infinite potential well, we can rewrite this as: \(-\frac{\hbar^2}{2m}(\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}) = E\Psi\) **Step 2: Solve the Schrödinger equation using separation of variables**

Using the method of separation of variables, we assume that the wave function can be written as a product of three single-variable functions: \(\Psi(x, y, z) = X(x)Y(y)Z(z)\) Substituting this into the equation above, we get: \(-\frac{\hbar^2}{2m}(\frac{1}{X}\frac{\partial^2 X}{\partial x^2} + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} + \frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}) = E\Psi\) Now, divide both sides by \(\Psi\), then separate it into three different equations, each of them representing components x, y, and z: \(\frac{-\frac{\hbar^2}{2m}(\frac{1}{X}\frac{\partial^2 X}{\partial x^2})}{X(x)Y(y)Z(z)} + \frac{-\frac{\hbar^2}{2m}(\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2})}{X(x)Y(y)Z(z)} + \frac{-\frac{\hbar^2}{2m}(\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2})}{X(x)Y(y)Z(z)} =\frac{E}{X(x)Y(y)Z(z)}\) Therefore, we have: \(-\frac{\hbar^2}{2m}\frac{1}{X}\frac{\partial^2 X}{\partial x^2} = E_x X\) \(-\frac{\hbar^2}{2m}\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2} = E_y Y\) \(-\frac{\hbar^2}{2m}\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2} = E_z Z\) Solving each equation, we find: \(X(x) = \sqrt{\frac{2}{L_x}}\sin(\frac{n_x\pi x}{L_x})\) \(Y(y) = \sqrt{\frac{2}{L_y}}\sin(\frac{n_y\pi y}{L_y})\) \(Z(z) = \sqrt{\frac{2}{L_z}}\sin(\frac{n_z\pi z}{L_z})\) Where \(n_x, n_y, n_z\) are positive integers (\(1, 2, 3,\)...) and the corresponding energies \(E_x, E_y\), and \(E_z\) are given by: \(E_x = \frac{n_x^2\pi^2\hbar^2}{2mL_x^2}\) \(E_y = \frac{n_y^2\pi^2\hbar^2}{2mL_y^2}\) \(E_z = \frac{n_z^2\pi^2\hbar^2}{2mL_z^2}\) Finally, finding the particle's wave function \(\Psi(x, y, z)\) and total energy \(E\):

The three-dimensional wave function: \(\Psi(x, y, z) = X(x)Y(y)Z(z) = \sqrt{\frac{8}{L_x L_y L_z}}\sin(\frac{n_x\pi x}{L_x})\sin(\frac{n_y\pi y}{L_y})\sin(\frac{n_z\pi z}{L_z})\) The total energy is the sum of \(E_x, E_y\), and \(E_z\): \(E = E_x + E_y + E_z = \frac{\pi^2\hbar^2}{2m}( \frac{n_x^2}{L_x^2}+ \frac{n_y^2}{L_y^2}+ \frac{n_z^2}{L_z^2})\) Now that we have the wave function and energy expressions, we can proceed with finding the lowest energy allowed for the electron. **Step 3: Find the lowest energy allowed**

The lowest energy state corresponds to the lowest possible values of \(n_x, n_y, n_z\), which are equal to 1. Therefore, let: \(n_x = n_y = n_z = 1\) Substitute the values of \(n_x\), \(n_y\) and \(n_z\) into the energy expression: \(E = \frac{\pi^2\hbar^2}{2m}( \frac{1}{L_x^2}+ \frac{1}{L_y^2}+ \frac{1}{L_z^2})\) Now, we can use the values of \(L_x = L_y = L_z = 10^{-10}\mathrm{~m}\), \(\hbar = 1.05457182 × 10^{-34} J⋅s\) (reduced Planck constant), and \(m = 9.10938356 × 10^{-31} kg\) (electron mass):

Insert the values to compute the lowest energy allowed: \(E = \frac{\pi^2(1.05457182 × 10^{-34})^2}{2(9.10938356 × 10^{-31})(10^{-10})^2} (3)\) Solving for E, we get: \(E \approx 1.76 \times 10^{-18} J\) Thus, the lowest energy allowed for the electron in this three-dimensional infinite potential well is approximately \(1.76 \times 10^{-18}\) Joules.