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Chapter 31: Problem 41

Give the value of the quantum number \(\ell,\) if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of (a) \(\sqrt{6}(h / 2 \pi),\) (b) \(\sqrt{15}(h / 2 \pi),\) (c) \(\sqrt{30}(h / 2 \pi)\), or (d) \(\sqrt{36}(h / 2 \pi)\).

Short Answer

Expert verified
(a) \( \ell = 2 \), (b) \( \ell = 3 \), (c) \( \ell = 5 \), (d) \( \ell = 6 \).

Step by step solution

01

Understanding Orbital Angular Momentum

In quantum mechanics, the magnitude of the orbital angular momentum for an electron in an atom is given by the formula: \( L = \sqrt{\ell(\ell+1)} \frac{h}{2\pi} \), where \( \ell \) is the azimuthal (or angular momentum) quantum number.
02

Matching Given Magnitudes to the Formula

We need to compare the given magnitudes to \( \sqrt{\ell(\ell+1)} \frac{h}{2\pi} \) and solve for \( \ell \).
03

Solve for (a) \( \sqrt{6}(h / 2\pi) \)

Given \( L = \sqrt{6} \frac{h}{2\pi} \), equate \( \sqrt{\ell(\ell+1)} \) to \( \sqrt{6} \).\( \ell(\ell+1) = 6 \). Solving the equation \( \ell^2 + \ell - 6 = 0 \) gives roots \( \ell = 2 \) and \( \ell = -3 \). Since \( \ell \) must be non-negative, \( \ell = 2 \).
04

Solve for (b) \( \sqrt{15}(h / 2\pi) \)

Given \( L = \sqrt{15} \frac{h}{2\pi} \), equate \( \sqrt{\ell(\ell+1)} \) to \( \sqrt{15} \). \( \ell(\ell+1) = 15 \). Solving the equation \( \ell^2 + \ell - 15 = 0 \) gives roots \( \ell = 3 \) and \( \ell = -5 \). Since \( \ell \) must be non-negative, \( \ell = 3 \).
05

Solve for (c) \( \sqrt{30}(h / 2\pi) \)

Given \( L = \sqrt{30} \frac{h}{2\pi} \), equate \( \sqrt{\ell(\ell+1)} \) to \( \sqrt{30} \). \( \ell(\ell+1) = 30 \). Solving the equation \( \ell^2 + \ell - 30 = 0 \) gives roots \( \ell = 5 \) and \( \ell = -6 \). Since \( \ell \) must be non-negative, \( \ell = 5 \).
06

Solve for (d) \( \sqrt{36}(h / 2\pi) \)

Given \( L = \sqrt{36} \frac{h}{2\pi} \), equate \( \sqrt{\ell(\ell+1)} \) to \( \sqrt{36} \). \( \ell(\ell+1) = 36 \). Solving the equation \( \ell^2 + \ell - 36 = 0 \) gives roots \( \ell = 6 \) and \( \ell = -6 \). Since \( \ell \) must be non-negative, \( \ell = 6 \).

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Key Concepts

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

Orbital Angular Momentum
In the world of quantum mechanics, orbital angular momentum plays a key role in understanding atomic structure. It suggests the momentum an electron possesses due to its orbital motion around the nucleus. The magnitude of this angular momentum is represented by the formula: \[ L = \sqrt{\ell(\ell+1)} \frac{h}{2\pi} \]where:
  • \( L \) is the orbital angular momentum
  • \( \ell \) is the azimuthal quantum number
  • \( h \) is Planck's constant
This equation highlights that the angular momentum is quantized, meaning it can only take specific, discrete values determined by the azimuthal quantum number \( \ell \). Understanding this concept allows us to link quantum numbers to physical properties of the electron's motion around the nucleus in an atom.
Azimuthal Quantum Number
The azimuthal quantum number, often denoted by \( \ell \), directly relates to the shape of the electron's orbital through its values. It specifies the subshells or sublevels of an electron within an atom. Here鈥檚 how it works:
  • The azimuthal quantum number \( \ell \) can have integer values starting from 0 up to \( n-1 \), where \( n \) is the principal quantum number or the main energy level of the electron.
  • Each value of \( \ell \) corresponds to a specific type of orbital: \( \ell = 0, 1, 2, 3 \) for s, p, d, and f orbitals respectively.
The value of \( \ell \) determines not just the shape, but also the number of different orbitals that compose a particular subshell. For instance, an \( \ell \) value of 2 suggests a "d" orbital, where electrons have particular energy and momentum characteristics. Understanding this quantum number is essential for comprehending the electron configuration and chemical behavior of elements.
Hydrogen Atom
The hydrogen atom serves as the fundamental model in the study of quantum mechanics due to its simplicity. It consists of a single electron orbiting a single proton nucleus. Despite its simplicity, the hydrogen atom exhibits all the complex behaviors governed by quantum mechanics principles. Here鈥檚 why it鈥檚 important:
  • The hydrogen atom's energy levels can be exactly solved using quantum mechanics, providing key insights into the principles of electron configuration and spectral lines.
  • Its sole electron allows easy illustration and understanding of how quantum numbers, like the azimuthal quantum number \( \ell \), influence atomic structure and states.
Studying the hydrogen atom thus sets the foundation for understanding more complex atoms and molecules. It鈥檚 the perfect starting point for diving into the world of quantum mechanics, offering a clear view of how quantum numbers dictate physical and chemical properties of elements.

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Most popular questions from this chapter

Give the electronic configuration for the ground state of carbon.

(a) Find the radius of the \(n=4\) Bohr orbit of a doubly ionized lithium atom \(\left(\mathrm{Li}^{2+}, \mathrm{Z}=3\right)\). (b) Is the energy required to raise an electron from the \(n=4\) state to the \(n=5\) state in \(\mathrm{Li}^{2+}\) greater than, less than, or equal to the energy required to raise an electron in hydrogen from the \(n=4\) state to the \(n=5\) state? Explain. (c) Verify your answer to part (b) by calculating the relevant energies.

A hydrogen atom has an orbital angular momentum with a magnitude of \(10 \sqrt{57}(h / 2 \pi) .\) (a) Determine the value of the quantum number \(\ell\) for this atom. (b) What is the minimum possible value of this atom's principal quantum number, \(n ?\) Explain. (c) If \(10 \sqrt{57}(h / 2 \pi)\) is the maximum orbital angular momentum this a tom can have, what is its energy?

A hydrogen atom is in the initial state \(n_{i}=n,\) where \(n>1\) (a) Find the frequency of the photon that is emitted when the electron jumps to state \(n_{f}=n-1 .\) (b) Find the frequency of the electron's orbital motion in the state \(n\). (c) Compare your results for parts (a) and (b) in the limit of large \(n\).

Laser Eye Surgery In laser eye surgery, the laser emits a 1.45 -ns pulse focused on a spot that is \(34.0 \mu \mathrm{m}\) in diameter. (a) If the energy contained in the pulse is \(2.75 \mathrm{mJ},\) what is the power per square meter (the irradiance) associated with this beam? (b) Suppose a molecule with a diameter of \(0.650 \mathrm{nm}\) is irradiated by the laser beam. How much energy does the molecule receive in one pulse from the laser? (The energy obtained in part (b) is more than enough to dissociate a molecule.)
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