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Chapter 30: Problem 28

The electron in a certain hydrogen atom has an angular momentum of \(8.948 \times 10^{-34} \mathrm{J} \cdot \mathrm{s} .\) What is the largest possible magnitude for the \(z\) component of the angular momentum of this electron? For accuracy, use \(h=6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\).

Short Answer

Expert verified
The largest possible magnitude for the \( z \)-component of the angular momentum is \( 8.432 \times 10^{-34} \) J路s.

Step by step solution

01

Understand Angular Momentum

Angular momentum of an electron in a hydrogen atom is quantized and calculated using the formula: \( L = \sqrt{l(l+1)} \cdot \hbar \). Here, \( \hbar = \frac{h}{2\pi} \) and \( l \) is an integer quantum number.
02

Calculate Reduced Planck's Constant

Calculate the reduced Planck's constant using \( \hbar = \frac{h}{2\pi} \). Use \( h = 6.626 \times 10^{-34} \) J路s. Hence, \( \hbar = \frac{6.626 \times 10^{-34}}{2 \cdot 3.1416} = 1.054 \times 10^{-34} \) J路s.
03

Set Up the Equation

The given angular momentum \( (L) \) is \( 8.948 \times 10^{-34} \) J路s. So the equation is \( \sqrt{l(l+1)} \cdot \hbar = 8.948 \times 10^{-34} \). Substituting \( \hbar \), we get \( \sqrt{l(l+1)} \times 1.054 \times 10^{-34} = 8.948 \times 10^{-34} \).
04

Solve for Quantum Number \( l \)

Divide both sides by \( 1.054 \times 10^{-34} \) to isolate \( \sqrt{l(l+1)} \), so \( \sqrt{l(l+1)} = 8.488 \). Solve \( l(l+1) = (8.488)^2 = 72.02 \). Find the integer \( l \) such that \( l(l+1) \approx 72.02 \). Try \( l = 8 \), giving \( 8 \cdot 9 = 72 \).
05

Calculate Maximum \( z \) Component of Angular Momentum

The maximum \( z \)-component of angular momentum is given by \( L_z = m_l \cdot \hbar \), where \( m_l \) ranges from \(-l\) to \(l\). For maximum \( L_z \), \( m_l = l = 8 \). Thus, \( L_z = 8 \cdot 1.054 \times 10^{-34} = 8.432 \times 10^{-34} \) J路s.

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Key Concepts

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

Quantum Number
In the study of atomic physics, quantum numbers are vital in describing the properties of electrons in atoms. Each electron in an atom is characterized by a set of quantum numbers, which give information about its energy level, angular momentum, magnetic orientation, and spin.

  • Principal Quantum Number (\( n \)): Indicates the energy level and the relative size of the electron cloud. The larger the value of \( n \), the further the electron is from the nucleus.
  • Azimuthal Quantum Number (\( l \)): Associated with the shape of the electron's orbital and relates directly to its angular momentum. \( l \) is an integer ranging from 0 to \( n-1 \). For instance, if \( l = 8 \), this suggests that the orbital has a higher angular momentum.
  • Magnetic Quantum Number (\( m_l \)): Specifies the orientation of the orbital in space, ranging from \( -l \) to \( +l \).
  • Spin Quantum Number (\( m_s \)): Refers to the direction of the electron's spin, with possible values of \( +\frac{1}{2} \) or \( -\frac{1}{2} \).
By understanding these quantum numbers, you can predict an electron's behavior and its role within an atom's structure. In this exercise, understanding the azimuthal quantum number \( l \) was essential to solve for the electron's angular momentum.
Hydrogen Atom
The hydrogen atom is a well-studied model in quantum mechanics, primarily because it has a single electron orbiting its nucleus. This simplicity makes it an excellent candidate for investigating fundamental principles of atomic physics and quantum mechanics.

  • The single electron in a hydrogen atom moves in a spherical orbit around the nucleus, which contains just one proton.
  • Since hydrogen has only one electron, the interactions that occur are straightforward and provide clear insights into electron behavior.
  • Studying these atoms allows us to understand how quantization affects electron orbits, angular momentum, and energy levels.
In this exercise, the hydrogen atom's angular momentum is defined using quantized values, emphasizing its role as an ideal model for exploring quantum mechanical concepts.
Planck's Constant
Planck's constant is a fundamental value used throughout quantum mechanics, represented by the symbol \( h \). It is essential for describing the quantization of energy levels in atoms.

Key Points:

  • Planck's constant (\( h = 6.626 \times 10^{-34} \) J路s) is crucial in formulas that determine an electron's behavior, energy transition, and angular momentum.
  • The reduced Planck's constant (\( \hbar \)) is derived using \( \hbar = \frac{h}{2\pi} \). This constant is frequently used when dealing with angular momentum and rotation.
  • In this particular exercise, \( \hbar \) is a vital component in calculating the maximum \( z \)-component of an electron's angular momentum. Using these constants helped determine the quantized values for angular momentum and influence quantum number findings.
Planck's constant underscores the discrete nature of energy and momentum at the quantum level, guiding the calculations and solving steps for understanding the behavior of electrons in atoms like hydrogen.

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Most popular questions from this chapter

A singly ionized helium atom \(\left(\mathrm{He}^{+}\right)\) has only one electron in orbit about the nucleus. What is the radius of the ion when it is in the second excited state?

A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the \(n=3\) excited state. The atom returns to the ground state by emitting two photons. What are their wavelengths?

The maximum value for the magnetic quantum number in state A is \(m_{\ell}=2,\) while in state \(\mathrm{B}\) it is \(m_{\ell}=1 .\) What is the ratio \(L_{\mathrm{A}} / L_{\mathrm{B}}\) of the magnitudes of the orbital angular momenta of an electron in these two states?

When an electron makes a transition between energy levels of an atom, there are no restrictions on the initial and final values of the principal quantum number \(n .\) According to quantum mechanics, however, there is a rule that restricts the initial and final values of the orbital quantum number \(\ell\). This rule is called a selection rule and states that \(\Delta \ell=\pm 1\). In other words, when an electron makes a transition between energy levels, the value of \(\ell\) can only increase or decrease by one. The value of \(\ell\) may not remain the same nor may it increase or decrease by more than one. According to this rule, which of the following energy level transitions are allowed? (a) \(2 \mathrm{s} \rightarrow 1 \mathrm{s}\) (b) \(2 p \rightarrow 1 s\) (c) \(4 p \rightarrow 2 p(d) 4 s \rightarrow 2 p\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

In the ground state, the outermost shell \((n=1)\) of helium (He) is filled with electrons, as is the outermost shell \((n=2)\) of neon (Ne). The full outermost shells of these two elements distinguish them as the first two so-called "noble gases." Suppose that the spin quantum number \(m_{\mathrm{s}}\) had three possible values, rather than two. If that were the case, which elements would be (a) the first and (b) the second noble gases? Assume that the possible values for the other three quantum numbers are unchanged, and that the Pauli exclusion principle still applies.
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