Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An electron is in a three-dimensional box with side lengths \(L_X =\) 0.600 nm and \(L_Y = L_Z = 2L_X\). What are the quantum numbers \(n_X, n_Y,\) and \(n_Z\) and the energies, in eV, for the four lowest energy levels? What is the degeneracy of each (including the degeneracy due to spin)?

Short Answer

Expert verified
Quantum numbers for lowest levels are: (1,1,1), (1,1,2), (1,2,1), (2,1,1). Energies are computed using energy formula. Degeneracies include spin.

Step by step solution

01

Understand the Problem

We have an electron in a 3D box with sides \( L_X = 0.600 \text{ nm} \) and \( L_Y = L_Z = 2L_X \). We need to find possible quantum numbers \( n_X, n_Y, \) and \( n_Z \) for the four lowest energies and also calculate these energies.
02

Write the Formula for Energy Levels in a 3D Box

The energy levels for an electron in a 3D box are given by the formula: \[ E_{n_x,n_y,n_z} = \left(\frac{h^2}{8m}\right) \left(\frac{n_x^2}{L_X^2} + \frac{n_y^2}{L_Y^2} + \frac{n_z^2}{L_Z^2}\right) \] where \( h \) is Planck's constant, \( m \) is the electron mass, and \( n_X, n_Y, n_Z \) are the quantum numbers.
03

Substitute Values into the Energy Formula

Convert \( L_Y \) and \( L_Z \) to lengths in nm: \( L_Y = L_Z = 1.2 \text{ nm} \). Substitute \( L_X = 0.600 \text{ nm} \) and \( L_Y = L_Z = 1.2 \text{ nm} \) into the energy formula.
04

Determine Quantum Numbers for the Four Lowest Energy Levels

For the lowest energy states, try combinations of \( n_X, n_Y, \) and \( n_Z \) where the sum of their squares divided by the respective box lengths yields the smallest values. The lowest four combinations are (1,1,1), (1,1,2), (1,2,1), and (2,1,1).
05

Calculate the Energies

Compute the energies using the formula. For the first combination (1,1,1), compute as follows: \[ E_{1,1,1} = \left(\frac{h^2}{8m}\right) \left(\frac{1^2}{(0.600 \text{ nm})^2} + \frac{1^2}{(1.2 \text{ nm})^2} + \frac{1^2}{(1.2 \text{ nm})^2} \right) \] Repeat for the other combinations (for \( n_X, n_Y, n_Z\) as (1,1,2), (1,2,1), (2,1,1)). Calculate energies in eV.
06

Calculate Degeneracy Including Spin

Since the electron can have spin up or spin down, there is a degeneracy of 2 for each non-degenerate state. For states with identical energy levels like (1,2,1), (2,1,1), and (1,1,2), add their degeneracies. Calculate total degeneracy for each level.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Three-dimensional box
A three-dimensional box in quantum mechanics is a simple theoretical model used to understand the behavior of particles like electrons confined in a compact space. Imagine having a box where an electron moves freely within certain boundaries. These boundaries in our case are defined by the side lengths function:
  • Two sides, \(L_Y\) and \(L_Z\), are doubled in length compared to one side, \(L_X\), implying the box isn't cubical.
  • In our specific example, \(L_X\) is 0.600 nm, meaning \(L_Y\) and \(L_Z\) are each 1.2 nm (since \(L_Y = L_Z = 2L_X\)).
This box concept helps us apply boundary conditions and use the shrinked dimensions to calculate potential energy levels of electrons within these confines. Understanding this basic setup is essential before delving into how particles like electrons behave under these conditions.
Quantum numbers
Quantum numbers are essential tools in quantum mechanics. They help us pinpoint the allowed energy levels for particles within confined systems like our three-dimensional box. Each quantum number represents one dimension of the box and how the particle's position or state is quantized within that dimension.
  • \(n_X\), \(n_Y\), and \(n_Z\) represent quantum numbers in each of the three dimensions respectively.
  • These numbers are integers starting from 1 (not zero). They designate the node counts in standing wave patterns in the box.
These values are pivotal in determining the energy of the particle since they indicate how the electron "bounces" around inside the box. For each energy level, the quantum numbers are sought in combinations where their squared sum divided by their respective box dimension gives the lowest possible values. This principle ensures we can identify the lowest energy states for our electron.
Energy levels
Energy levels in a quantum box are derived based on a mathematical model. The energy at each level can be mathematically described as: \[E_{n_X,n_Y,n_Z} = \left(\frac{h^2}{8m}\right) \left(\frac{n_X^2}{L_X^2} + \frac{n_Y^2}{L_Y^2} + \frac{n_Z^2}{L_Z^2}\right)\]where:
  • \(E_{n_X,n_Y,n_Z}\) represents the energy corresponding to quantum numbers \(n_X, n_Y,\) and \(n_Z\).
  • \(h\) is Planck's constant and \(m\) is the mass of the electron.
For our particular case, we seek energy combinations that are the lowest, based on quantum number permutations. Once the permitted combinations, like (1,1,1), (1,1,2), etc., are known, energies can be calculated by substituting these into the formula.Additionally, since energy depends on squared quantum numbers, even a small change in these numbers can lead to significant difference in energy, illustrating why their correct assignment is vital in quantum level calculations.
Electron spin
Electron spin is an intrinsic form of angular momentum carried by electrons. It can have one of two possible values:
  • Spin-up (+1/2)
  • Spin-down (-1/2)
This property affects the degeneracy of energy-level states in quantum systems. Degeneracy refers to the number of different states that share the same energy level. In a box model, each energy level initially found would theoretically only allow one state. However, when accounting for electrons' spins:
  • Each energy state is doubled in possibility due to the two independent spin states, leading to an effective degeneracy of 2.
For systems where certain energy levels coincide for different quantum number sets, the degeneracy increases. This influence of spin is crucial to incorporate in calculating accurate levels of energy states for a more comprehensive understanding.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron is in the hydrogen atom with \(n\) = 5. (a) Find the possible values of \(L\) and \(L$$_z\) for this electron, in units of \(\hslash\). (b) For each value of \(L\), find all the possible angles between \(\vec{L}\) and the z-axis. (c) What are the maximum and minimum values of the magnitude of the angle between \(L\) S and the z-axis?

The normalized radial wave function for the \(2p\) state of the hydrogen atom is \(R_2{_p}\) = \(( 1/ \sqrt{24a^5}\))\(re$$^-{^r}{^/}{^2}{^a}\). After we average over the angular variables, the radial probability function becomes \(P$$(r)\) \(dr\) = \((R_2{_p}$$)^2\)r\(^2\) \(dr\). At what value of \(r\) is \(P$$(r)\) for the \(2p\) state a maximum? Compare your results to the radius of the \(n\) = 2 state in the Bohr model.

(a) Show that the total number of atomic states (including different spin states) in a shell of principal quantum number \(n\) is \(2n^2\). \([Hint\): The sum of the first \(N\) integers 1 + 2 + 3 + \(\cdots\) + \(N\) is equal to \(N$$(N + 1)\)/2.] (b) Which shell has 50 states?

A particle is in the three-dimensional cubical box of Section 41.2. (a) Consider the cubical volume defined by \(0 \leq x \leq L/4, 0 \leq y \leq L/4\), and \(0 \leq z \leq L/4\). What fraction of the total volume of the box is this cubical volume? (b) If the particle is in the ground state \((n_X = 1, n_Y = 1, n_Z = 1)\), calculate the probability that the particle will be found in the cubical volume defined in part (a). (c) Repeat the calculation of part (b) when the particle is in the state \(n_X = 2, n_Y = 1, n_Z = 1\).

A hydrogen atom in a 3\(p\) state is placed in a uniform external magnetic field \(\vec B\). Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. (a) What field magnitude \(B\) is required to split the 3\(p\) state into multiple levels with an energy difference of 2.71 \(\times\) 10\(^{-5}\) eV between adjacent levels? (b) How many levels will there be?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free