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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

Expert verified
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

An experimental measurement of the energy levels of a hydrogen molecule, \(\mathrm{H}_{2}\), shows that the energy levels are evenly spaced and separated by about \(9 \cdot 10^{-20} \mathrm{~J}\). A reasonable model of one of the hydrogen atoms would then seem to be that of a hydrogen atom in a simple harmonic oscillator potential. Assuming that the hydrogen atom is attached by a spring with a spring constant \(k\) to the molecule, what is the spring constant \(k\) ?

A beam of electrons moving in the positive \(x\) -direction encounters a potential barrier that is \(2.51 \mathrm{eV}\) high and \(1.00 \mathrm{nm}\) wide. Each electron has a kinetic energy of \(2.50 \mathrm{eV},\) and the electrons arrive at the barrier at a rate of 1000 electrons/s (1000. electrons every second). What is the rate \(\mathrm{I}_{\mathrm{T}}\) in electrons/s at which electrons pass through the barrier, on average? What is the rate \(\mathrm{I}_{\mathrm{R}}\) in electrons/s at which electrons reflect back from the barrier, on average? Determine and compare the wavelengths of the electrons before and after they pass through the barrier.

An electron is confined in a one-dimensional infinite potential well of \(1.0 \mathrm{nm}\). Calculate the energy difference between a) the second excited state and the ground state, and b) the wavelength of light emitted by this radiative transition.

Find the probability of finding an electron trapped in a one-dimensional infinite well of width \(2.00 \mathrm{nm}\) in the \(n=2\) state between 0.800 and \(0.900 \mathrm{nm}\) (assume that the left edge of the well is at \(x=0\) and the right edge is at \(x=2.00 \mathrm{nm}\) ).

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