An electron is confined in a three-dimensional cubic space of \(L^{3}\) with infinite potentials. a) Write down the normalized solution of the wave function in the ground state. b) How many energy states are available up to the second excited state from the ground state? (Take the electron spin into account.)

Short Answer

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Answer: The normalized wave function for the ground state is \(\psi_{111}(x,y,z) = \sqrt{\frac{8}{L^3}}\sin(\frac{\pi x}{L})\sin(\frac{\pi y}{L})\sin(\frac{\pi z}{L})\). The total number of energy states available up to the second excited state, considering the electron's spin, is 13.

Step by step solution

01

Solve the Schrödinger equation for the cubic box#

In this step, we will solve the time-independent Schrödinger equation for the 3D cubic box. The equation for the energy eigenfunction is: \((-\frac{\hbar^2}{2m}\nabla^2 + V(r))\psi(r) = E\psi(r) \) Here, \(\nabla^2\) is the Laplacian, \(\hbar\) is the reduced Planck's constant, \(m\) is the mass of the electron, \(V(r)\) is the potential, and \(E\) is the energy. Since the potential is zero inside the box and infinite outside, we can write this as: \(-\frac{\hbar^2}{2m}\nabla^2 \psi(r) = E\psi(r) \) Breaking down the Laplacian to cartesian coordinates gives: \(-\frac{\hbar^2}{2m}(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})\psi(x,y,z) = E\psi(x,y,z) \) Solving this equation using separation of variables, lets us write the wave function as the product of three separate functions: \(\psi(x,y,z) = X(x)Y(y)Z(z) \)
02

Find the individual wave functions and normalization constant#

After solving the separated Schrödinger equation for X, Y, and Z, we obtain: \(X(x) = \sqrt{\frac{2}{L}}\sin(\frac{n_x \pi x}{L})\) \(Y(y) = \sqrt{\frac{2}{L}}\sin(\frac{n_y \pi y}{L})\) \(Z(z) = \sqrt{\frac{2}{L}}\sin(\frac{n_z \pi z}{L})\) Here, \(n_x\), \(n_y\), and \(n_z\) are the quantum numbers which cannot be zero. The ground state corresponds to the minimum energy, which occurs when \(n_x = n_y = n_z = 1\). The overall wave function for the ground state is: \(\psi_{111}(x,y,z) = \sqrt{\frac{8}{L^3}}\sin(\frac{\pi x}{L})\sin(\frac{\pi y}{L})\sin(\frac{\pi z}{L})\) This is the normalized wave function for the ground state.
03

Determine the available energy states up to the second excited state#

To determine the available energy states, we need to find the allowed values of \(n_x\), \(n_y\), and \(n_z\) for each state. The energy of the eigenstate is given by: \(E_{n_x n_y n_z} = \frac{\hbar^2 \pi^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2)\) Since the ground state corresponds to \(n_x = n_y = n_z = 1\), the first excited state corresponds to any combination that comes next in ascending order. Here, the possible energy levels are: - \(n_x = 1, n_y = 1, n_z = 2\) (Energy Level 1) - \(n_x = 1, n_y = 2, n_z = 1\) (Energy Level 1) - \(n_x = 2, n_y = 1, n_z = 1\) (Energy Level 1) - \(n_x = 1, n_y = 2, n_z = 2\) (Energy Level 2) - \(n_x = 2, n_y = 1, n_z = 2\) (Energy Level 2) - \(n_x = 2, n_y = 2, n_z = 1\) (Energy Level 2) Considering the electron spin, we have two spin states for each energy level (\(\text{spin}_z = \frac{1}{2}\) and \(\text{spin}_z = -\frac{1}{2}\)). Therefore, we need to multiply the number of energy levels by two to account for the electron's spin. The total number of energy states available up to the second excited state is: \(1 (\text{ground state}) + 3 \cdot 2 (\text{first excited state}) + 3 \cdot 2 (\text{second excited state}) = 1 + 6 + 6 = 13\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Schrödinger Equation
The Schrödinger equation is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes over time. In the textbook exercise, we explore a simplified version known as the time-independent Schrödinger equation, which is used to calculate the energy levels of a system that does not change with time.

The equation takes the form:
\[ (-\frac{\hbar^2}{2m}abla^2 + V(r))\psi(r) = E\psi(r) \]
where:\begin{itemize}\item \( \hbar \) represents the reduced Planck's constant, indicating the scale of quantum effects. \item \( m \) is the mass of the particle, such as an electron. \item \( abla^2 \) is the Laplacian operator, describing how the wave function curves in space. \item \( V(r) \) is the potential energy as a function of position. \item \( \psi(r) \) is the quantum wave function. \item \( E \) is the energy of the quantum state.\end{itemize}
In this specific exercise, the equation simplifies due to the infinite potential outside of a 'cubic box,' leaving us with \( -\frac{\hbar^2}{2m}abla^2 \psi(r) = E\psi(r) \). Here, we are dealing with a particle trapped in a three-dimensional box, a classic case in quantum mechanics that illustrates discretized energy levels and particle confinement.
Quantum Wave Function
The quantum wave function, denoted as \( \psi \), is a mathematical function that describes the quantum state of a particle or system. In the context of our problem, the wave function represents an electron confined in a cubic box.

For the ground state of a particle in a three-dimensional box, the normalized wave function is:
\[ \psi_{111}(x,y,z) = \sqrt{\frac{8}{L^3}}\sin(\frac{\pi x}{L})\sin(\frac{\pi y}{L})\sin(\frac{\pi z}{L}) \]
The quantum wave function has several key characteristics:
  • It contains all the information about the system.
  • Its square magnitude \(|\psi|^2\) gives the probability density of finding the particle at a particular location.
  • It must be normalized so that the total probability of finding the particle within the entire space is 1.

The exercise guides us through the process of calculating and normalizing the wave function for an electron in a 'quantum box.' This concept cements the probabilistic nature of quantum mechanics, differing from classical mechanics, where objects have definite positions and velocities.
Quantum Energy States
Quantum energy states reflect the discrete energy levels available to a particle confined in a quantum system, such as an electron in a box. These states are a direct consequence of the wave-like nature of particles in quantum mechanics, leading to quantization of physical properties.

The energy of a quantum state in our cubic box problem is given by:
\[ E_{n_x n_y n_z} = \frac{\hbar^2 \pi^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2) \]
This expression shows that the energy levels depend on the quantum numbers (\( n_x, n_y, n_z \)), which represent the states of motion along the respective axis of the box. The ground state has the lowest energy, and higher energy states correspond to higher quantum numbers.

When considering electron spin, each energy level can host two electrons due to the Pauli exclusion principle—one with spin-up and one with spin-down. Therefore, for each energy level derived from the quantum numbers, we can have two distinct states due to spin. In the example, the increment from the ground state to the first and second excited states is demonstrated, revealing the count of available energy states for an electron in a three-dimensional box up to the second excited state, which totals to 13 including spin considerations.

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Most popular questions from this chapter

Two long, straight wires that lie along the same line have a separation at their tips of \(2.00 \mathrm{nm}\). The potential energy of an electron in the gap is about \(1.00 \mathrm{eV}\) higher than it is in the conduction band of the two wires. Conduction-band electrons have enough energy to contribute to the current flowing in the wire. What is the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap?

Consider a water vapor molecule in a room \(4.00 \mathrm{~m} \times\) \(10.0 \mathrm{~m} \times 10.0 \mathrm{~m}\). a) What is the ground state energy of this molecule, treating it as a simple particle in a box? b) Compare this energy to the average thermal energy of such a molecule, taking the temperature to be \(300 . \mathrm{K}\). c) What can you conclude from the two numbers you just calculated?

State whether each of the following statements is true or false a) In a one-dimensional quantum harmonic oscillator, the energy levels are evenly spaced. b) In an infinite one-dimensional potential well, the energy levels are evenly spaced. c) The minimum total energy possible for a classical harmonic oscillator is zero. d) The correspondence principle states that because the minimum possible total energy for the classical simple harmonic oscillator is zero, the expected value for the fundamental state \((n=0)\) of the one-dimensional quantum harmonic oscillator should also be zero. e) The \(n=0\) state of the one-dimensional quantum harmonic oscillator is the state with the minimum possible uncertainty \(\Delta x \Delta p\)

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.

37.2 In an infinite square well, for which of the following states will the particle never be found in the exact center of the well? a) the ground state b) the first excited state c) the second excited state d) any of the above e) none of the above

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