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Chapter 30: Problem 27

An electron is in the \(n=5\) state. What is the smallest possible value for the angle between the \(z\) component of the orbital angular momentum and the orbital angular momentum?

Short Answer

Expert verified
The smallest angle is approximately 26.57 degrees.

Step by step solution

01

Understand Quantum Numbers

The electron is in an energy level described by the principal quantum number, which is given as \( n = 5 \). For any electron, the orbital angular momentum quantum number \( l \) can take integer values from 0 to \( n-1 \). Thus, \( l \) can be 0, 1, 2, 3, or 4 in this case.
02

Determine Magnetic Quantum Number

The magnetic quantum number \( m_l \) describes the orientation of the angular momentum and can take values ranging from \( -l \) to \( +l \). For each \( l \), \( m_l \) can be \(-l, -l+1, ..., 0, ..., l-1, l\).
03

Calculate Orbital Angular Momentum

The magnitude of the orbital angular momentum \( L \) is given by the formula \( L = \sqrt{l(l+1)} \hbar \), where \( \hbar \) is the reduced Planck constant.
04

Calculate the z-component of Orbital Angular Momentum

The z-component of the orbital angular momentum \( L_z \) is given by \( L_z = m_l \hbar \). For minimum angle, \(|m_l|\) should be as large as possible, so \( m_l = l \).
05

Calculate the Smallest Possible Angle

The smallest possible angle \( \theta \) between the z-component and the orbital angular momentum can be calculated using the formula \( \cos \theta = \frac{L_z}{L} \). Substituting \( L_z = l \hbar \) and \( L = \sqrt{l(l+1)} \hbar \), we find \( \cos \theta = \frac{l}{\sqrt{l(l+1)}} \).
06

Evaluate for the Largest l Value

Substitute \( l = 4 \) (since that's the largest value when \( n=5 \)), find \( \cos \theta = \frac{4}{\sqrt{4 \, (4+1)}} = \frac{4}{\sqrt{20}} = \frac{4}{2 \sqrt{5}} = \frac{2}{\sqrt{5}} \). Calculate \( \theta \) using \( \theta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 26.57^\circ \).

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Key Concepts

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

Quantum Numbers
In the world of quantum mechanics, quantum numbers are like unique identifiers for electrons. Each electron in an atom is specified by a set of four quantum numbers, which describe the electron's position and movement. These are:
  • The principal quantum number ( ) tells us which energy level the electron is in. It can be any positive integer (1, 2, 3, ...). The higher the number, the farther the electron is from the nucleus and the more energy it has.
  • The orbital angular momentum quantum number ( l ) is related to the shape of the electron's orbital, taking integer values from 0 to -1.
  • The magnetic quantum number ( ml ) indicates the orientation of the orbital in space and can range from -l to +l , including zero.
  • The spin quantum number describes the intrinsic spin of the electron, having values of either +1/2 or -1/2.
Each of these numbers plays a vital role in defining the behavior and interactions of electrons in atoms.
Orbital Angular Momentum
Orbital angular momentum (L_n) is a key concept in understanding how electrons move within an atom. It's akin to a tiny spinning top defining the electron's rotational motion around the nucleus.
The orbital angular momentum depends on the quantum number ldefined earlier, and its magnitude is given by:\[ L = \sqrt{l(l+1)} \hbar\]where \( \hbar \) is the reduced Planck constant.
This equation signifies that not all angular momentum values are possible, only quantized ones as defined by l.This quantization makes understanding electron motion slightly more complex than classical models.
Magnetic Quantum Number
The magnetic quantum number (ml) adds another layer to our understanding of electrons in atoms by describing an electron's orbital orientation in a magnetic field. It can take on integer values from -l to +l, inclusive.

For example, if \( l = 2 \), then \( m_l \) can be -2, -1, 0, 1, or 2.
  • The value of \( m_l \) affects the z-component of the orbital angular momentum, \( L_z \), as it's calculated by:\[ L_z = m_l \hbar\]
  • In a magnetic field, different orientations will have different energy levels due to their varying interactions with the field. This is known as the Zeeman effect.
The magnetic quantum number's role is pivotal when calculating how electrons are arranged in atoms under magnetic influences, contributing significantly to their magnetic properties.
Principal Quantum Number
The principal quantum number (), denoted by the symbol , is essentially indicative of the electron's energy level in an atom. Starting from =1, each subsequent integer represents a progressively higher energy level.

This number not only tells you the electron's energy but also gives you insight into its distance from the nucleus:
  • Higher values mean higher energy electrons that are further from the nucleus.
  • This pattern of energy levels helps define the various shells of electrons in an atom.
  • The total number of possible orbitals in a given energy level is \( n^2 \)
  • The principal quantum number sets the stage for determining the other three quantum numbers.
In many ways, it acts like a postal code, summarizing where an electron can "live" within the atom, further influencing not only its energy but also its reactivity and participation in chemical bonds.

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Most popular questions from this chapter

The \(K_{\beta}\) characteristic X-ray line for tungsten has a wavelength of \(1.84 \times 10^{-11} \mathrm{~m} .\) What is the difference in energy between the two energy levels that give rise to this line? Express the answer in (a) joules and (b) electron volts

Consider a particle of mass \(m\) that can exist only between \(x=0 \mathrm{~m}\) and \(x=+L\) on the \(x\) axis. We could say that this particle is confined to a "box" of length \(L\). In this situation, imagine the standing de Broglie waves that can fit into the box. For example, the drawing shows the first three possibilities. Note in this picture that there are either one, two, or three half-wavelengths that fit into the distance \(L\). Use Equation \(29.8\) for the de Broglie wavelength of a particle and derive an expression for the allowed energies (only kinetic energy) that the particle can have. This expression involves \(m, L\), Planck's constant, and a quantum number \(n\) that can have only the values \(1,2,3, \ldots .\)

The principal quantum number for an electron in an atom is \(n=6,\) and the magnetic quantum number is \(m_{\ell}=2 .\) What possible values for the orbital quantum number \(\ell\) could this electron have?

A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is \(514 \mathrm{~nm}\), and the power is \(1.5 \mathrm{~W}\). During surgery, the laser beam is turned on for \(0.050 \mathrm{~s}\). During this time, how many photons are emitted by the laser?

Suppose that the molybdenum \((Z=42)\) target in an X-ray tube is replaced by a silver \((Z=47)\) target. Do (a) the cutoff wavelength \(\lambda_{0}\) and (b) the wavelength of the \(K_{\alpha}\) X-ray photon increase, decrease, or remain the same? Assume that the voltage across the tube is constant and is sufficient to produce characteristic X-rays from both targets. Provide a reason for each answer. The voltage across the X-ray tube is \(35.0 \mathrm{kV}\). Determine (a) the cutoff wavelength \(\lambda_{0}\) and (b) the wavelengths of the \(K_{\alpha}\) X-ray photons emitted by the molybdenum and silver targets. Verify that your answers are consistent with your answers to the Concept Questions.
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