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Ch 41: Quantum Mechanics II: Atomic Structure
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 14th EditionCh 41: Quantum Mechanics II: Atomic StructureProblema 11
Capitolo 41, Problema 11

In a particular state of the hydrogen atom, the angle between the angular momentum vector L→\(\overrightarrow{L}\) and the zz-axis is u=26.6u = 26.6°. If this is the smallest angle for this particular value of the orbital quantum number ll, what is ll?

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Step 1: Recall the relationship between the orbital quantum number (l) and the angular momentum vector magnitude. The magnitude of the angular momentum vector is given by \( L = \sqrt{l(l+1)} \hbar \), where \( \hbar \) is the reduced Planck's constant.
Step 2: Understand the quantization of the projection of angular momentum along the z-axis. The projection \( L_z \) is given by \( L_z = m_l \hbar \), where \( m_l \) is the magnetic quantum number and can take integer values from \( -l \) to \( +l \).
Step 3: The angle \( \theta \) between the angular momentum vector \( \overrightarrow{L} \) and the z-axis is determined by \( \cos \theta = \frac{L_z}{L} \). Substitute \( L_z = m_l \hbar \) and \( L = \sqrt{l(l+1)} \hbar \) into this equation to get \( \cos \theta = \frac{m_l}{\sqrt{l(l+1)}} \).
Step 4: Since the problem states that \( \theta = 26.6^\circ \) is the smallest angle, \( \cos \theta \) must be maximized. The maximum value of \( \cos \theta \) occurs when \( m_l = l \), which corresponds to the smallest angle. Substitute \( \cos \theta = \frac{l}{\sqrt{l(l+1)}} \) and \( \theta = 26.6^\circ \) into the equation.
Step 5: Solve for \( l \) by rearranging the equation \( \cos 26.6^\circ = \frac{l}{\sqrt{l(l+1)}} \). Use trigonometric values for \( \cos 26.6^\circ \) and algebraic manipulation to isolate \( l \).

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Angular Momentum in Quantum Mechanics

In quantum mechanics, angular momentum is a fundamental property of particles, represented by the vector \( \overrightarrow{L} \). It is quantized, meaning it can only take on certain discrete values determined by the orbital quantum number \( l \). The magnitude of angular momentum is given by \( |\overrightarrow{L}| = \sqrt{l(l+1)}\hbar \), where \( \hbar \) is the reduced Planck's constant.
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Intro to Angular Momentum

Orbital Quantum Number (l)

The orbital quantum number \( l \) defines the shape of an electron's orbital and is integral to determining the angular momentum of an electron in an atom. It can take on integer values from 0 to \( n-1 \), where \( n \) is the principal quantum number. Each value of \( l \) corresponds to a specific type of orbital (s, p, d, f) and influences the angular distribution of the electron's probability density.
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Quantization of Angular Momentum

In quantum mechanics, the quantization of angular momentum implies that the angular momentum vector can only take specific orientations relative to an axis, such as the z-axis. The angle between the angular momentum vector and the z-axis is related to the magnetic quantum number \( m_l \), which can take values from \( -l \) to \( +l \). The smallest angle for a given \( l \) indicates the lowest energy state or configuration of the system.
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