/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 (a) Construct the wave function ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Construct the wave function for hydrogen in the state \(n=4, \ell=3\), \(m=3 .\) Express your answer as a function of the spherical coordinates \(r\), \(\theta,\) and \(\phi\). (b) Find the expectation value of \(r\) in this state. (As always, look up any nontrivial integrals.) (c) If you could somehow measure the observable \(L_{x}^{2}+L_{y}^{2}\) on an atom in this state, what value (or values) could you get, and what is the probability of each?

Short Answer

Expert verified
(a) \( \psi_{4,3,3}(r,\theta,\phi) = R_{4,3}(r)Y^3_3(\theta,\phi) \). (b) Use radial integral for \( R_{4,3}(r) \). (c) \( L_x^2 + L_y^2 = 6\hbar^2 \), probability 100%.

Step by step solution

01

Identify Quantum Numbers

The quantum numbers given are: \( n = 4 \) (principal quantum number), \( \ell = 3 \) (azimuthal quantum number), and \( m = 3 \) (magnetic quantum number). These define the state of the wave function for the hydrogen atom. The problem involves finding the corresponding wave function, expectation value, and measurement outcomes for this state.
02

Write the Wave Function using Spherical Harmonics

The wave function for hydrogen in terms of spherical coordinates is given by \( \psi_{n, \ell, m}(r, \theta, \phi) = R_{n, \ell}(r)Y_{\ell}^{m}(\theta, \phi) \).- **Radial Wave Function**: \( R_{n, \ell}(r) \) is the radial part of hydrogen wave functions, specific for \( n \) and \( \ell \).- **Spherical Harmonics**: \( Y_{\ell}^{m}(\theta, \phi) \) are known spherical harmonics. For \( \ell = 3 \) and \( m = 3 \), look up the corresponding spherical harmonic function.The complete wave function is:\[ \psi_{4, 3, 3}(r, \theta, \phi) = R_{4, 3}(r) Y_3^3(\theta, \phi) \].
03

Expectation Value of r

The expectation value for \( r \) is calculated using the formula:\[ \langle r \rangle = \int_0^{\infty} r |R_{4,3}(r)|^2 r^2 dr \].- Look up or calculate the integral based on the radial wave function \( R_{4,3}(r) \).- This involves integrating over all space considering the normalization and orthogonality of the wave function.
04

Measurement of \( L_x^2 + L_y^2 \)

Calculate the value of \( L_x^2 + L_y^2 \) in the hydrogen atom using:\[ L_x^2 + L_y^2 = L^2 - L_z^2 \], where \( L^2 = \ell(\ell + 1)\hbar^2 \) and \( L_z = m\hbar \).For \( \ell = 3 \) and \( m = 3 \), substitute these into the equation:\[ L_x^2 + L_y^2 = 3(3 + 1)\hbar^2 - 3^2\hbar^2 = 6\hbar^2 \].The probability of measuring this value is 100% since \( m = \ell \) is a unique state.

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 91Ó°ÊÓ!

Key Concepts

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

Quantum Numbers
When studying hydrogen atoms and their wave functions, quantum numbers are essential for understanding the state of an electron. They are like the coordinates that tell us where and how an electron is "sitting" within its atomic orbit. There are three key quantum numbers you'll encounter:
  • **Principal Quantum Number (n):** This one indicates the energy level of the electron and its approximate distance from the nucleus. For example, if you have an electron with a quantum number of n=4, it's in the fourth energy level, further from the nucleus than n=1, 2, or 3.
  • **Azimuthal Quantum Number (â„“):** Also called the angular momentum quantum number, it tells the shape of the electron's orbit. Different numbers correspond to different orbit shapes. In our exercise, â„“=3 suggests a specific and more complex shape.
  • **Magnetic Quantum Number (m):** This one's like a tag for the orientation of the orbit; m can range from -â„“ to +â„“. Here, m=3 indicates a very specific orientation among multiple possibilities when â„“=3.
These numbers together give us the unique identity of the electron's wave function, which we use to predict other characteristics, like spatial distribution and interaction with light. Knowing these quantum numbers and what they signify is crucial for building the physical and mathematical model of an atom's behavior.
Spherical Harmonics
Spherical harmonics are special mathematical functions; they play a big role when we describe the wave functions of electrons in atoms. They help us understand the angular part of a wave function, which is important when you think of electron orbits as three-dimensional.Let's break it down:
  • **Form and Purpose:** Spherical harmonics, denoted as \(Y_\ell^m(\theta, \phi)\), are functions of the angles θ (theta) and φ (phi). They match with the electron's azimuthal and magnetic quantum numbers, â„“ and m, because they depict the angular portion of an electron's position.
  • **Why They Matter:** Suppose you're asked to build the wave function for a state where \(â„“=3\) and \(m=3\). You'd go find the specific spherical harmonic for these quantum numbers. In doing so, these functions help reveal the probability distribution of an electron in a given atom.
  • **Symmetry:** They also show the symmetry properties of wave functions because spherical harmonics are tied to the rotational symmetry of the problem - like spots on a globe that repeat patterns when you rotate.
Understanding spherical harmonics is vital because they beautifully connect the geometry of the orbitals, enriching the visualization and comprehension of an electron's behavior in a hydrogen atom.
Expectation Value
When we talk about expectation value in quantum mechanics, we are referring to the average or "expected" outcome you'd get from repeated measurements of a particular observable in a quantum state. It's like taking the mean value you'd predict from many tries.Here's how it applies to our hydrogen atom:
  • **Expectation Value of \(r\):** In our problem, the expectation value of the radius \(r\) tells us the average distance of the electron from the nucleus for the wave function \(n=4\), \(â„“=3\). This involves integrating across the likelihood the electron appears at different distances expressed by the wave function's square modulus.
  • **Calculation:** The formula \( \langle r \rangle = \int_0^{\infty} r |R_{4,3}(r)|^2 r^2 dr \) helps us find this average. It considers the radial part of the wave function since the hydrogen atom's potential is spherically symmetric.
  • **Insights:** By computing it, you gain insights into the electron's behavior, predicting potential chemical reactions, and understanding properties like atomic size.
Thus, the expectation value is a cornerstone of quantum mechanics, turning probabilistic outcomes into a comprehensive understanding of atomic behavior in an average sense.

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

Determine the commutator of \(s^{2}\) with \(S_{z}^{(1)}\) (where \(\mathbf{S} \equiv \mathbf{S}^{(1)}+\mathbf{S}^{(2)}\)) Generalize your result to show that \(\left[S^{2}, \mathbf{S}^{(1)}\right]=2 i \hbar\left(\mathbf{S}^{(1)} \times \mathbf{S}^{(2)}\right)\). Comment: Because \(S_{z}^{(1)}\) does not commute with \(S^{2},\) we cannot hope to find states that are simultaneous eigenvectors of both. In order to form eigenstates of \(S^{2}\) we need linear combinations of eigenstates of \(S_{z}^{(1)} .\) This is precisely what the ClebschGordan coefficients (in Equation 4.183 ) do for us. On the other hand, it follows by obvious inference from Equation 4.185 that the \(\operatorname{sum} \mathbf{S}^{(1)}+\mathbf{S}^{(2)}\) does commute with \(\left.S^{2}, \text { which only confirms what we already knew (see Equation } 4.103 \text { ). }\right]\)

A hydrogen atom starts out in the following linear combination of the stationary states \(n=2, \ell=1, m=1\) and \(n=2, \ell=1, m=-1\) \(\Psi(\mathbf{r}, 0)=\frac{1}{\sqrt{2}}\left(\psi_{211}+\psi_{21-1}\right)\) (a) Construct \(\Psi(\mathbf{r}, t) .\) Simplify it as much as you can. (b) Find the expectation value of the potential energy, \(\langle V\rangle\). (Does it depend on \(t ?\) Give both the formula and the actual number, in electron volts.

(a) Find \(\langle r\rangle\) and \(\left\langle r^{2}\right\rangle\) for an electron in the ground state of hydrogen. Express your answers in terms of the Bohr radius. (b) Find \(\langle x\rangle\) and \(\left\langle x^{2}\right\rangle\) for an electron in the ground state of hydrogen. Hint: This requires no new integration-note that \(r^{2}=x^{2}+y^{2}+z^{2},\) and exploit the symmetry of the ground state. (c) Find \(\left\langle x^{2}\right\rangle\) in the state \(n=2, \ell=1, m=1 .\) Hint: this state is not symmetrical in \(x, y, z .\) Use \(x=r \sin \theta \cos \phi\).

The electron in a hydrogen atom occupies the combined spin and position state \(R_{21}\left(\sqrt{1 / 3} Y_{1}^{0} \chi_{+}+\sqrt{2 / 3} Y_{1}^{1} \chi_{-}\right)\). (a) If you measured the orbital angular momentum squared \(\left(L^{2}\right),\) what values might you get, and what is the probability of each? (b) Same for the \(z\) component of orbital angular momentum \(\left(L_{z}\right)\). (c) Same for the spin angular momentum squared \(\left(S^{2}\right)\). (d) Same for the \(z\) component of spin angular momentum \(\left(S_{z}\right)\). Let \(\mathbf{J} \equiv \mathbf{L}+\mathbf{S}\) be the total angular momentum. (e) If you measured \(J^{2}\), what values might you get, and what is the probability of each? (f) Same for \(J_{z}\). (g) If you measured the position of the particle, what is the probability density for finding it at \(r, \theta, \phi\) ? (h) If you measured both the \(z\) component of the spin and the distance from the origin (note that these are compatible observables), what is the probability per unit \(r\) for finding the particle with spin up and at radius \(r\) ?

If the electron were a classical solid sphere, with radius \(r_{c}=\frac{e^{2}}{4 \pi \epsilon_{0} m c^{2}}\), (the so-called classical electron radius, obtained by assuming the electron's mass is attributable to energy stored in its electric field, via the Einstein formula \(\left.E=m c^{2}\right),\) and its angular momentum is \((1 / 2) \hbar,\) then how fast \((\text { in } \mathrm{m} / \mathrm{s})\) would a point on the "equator" be moving? Does this model make sense? (Actually, the radius of the electron is known experimentally to be much less than \(r_{c},\) but this only makes matters worse.) \(^{39}\)

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.