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The orbital angular momentum of an electron has a magnitude of \(4.716 \times 10^{-34} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) . What is the angular- momentum quantum number \(/\) for this electron?

Short Answer

Expert verified
The angular-momentum quantum number \( l \) is 4.

Step by step solution

01

Understand the formula

The formula to calculate the magnitude of the angular momentum \(L\) in quantum mechanics is given by \(L = \sqrt{l(l+1)} \, \hbar\), where \(\hbar\) is the reduced Planck's constant \( (\hbar = 1.0545718 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}) \) and \( l \) is the angular-momentum quantum number.
02

Calculate \(\hbar\) term

Since \(L = 4.716 \times 10^{-34} \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} \), we can write \( \sqrt{l(l+1)} \hbar = 4.716 \times 10^{-34} \). Divide both sides by \(\hbar\): \[ \sqrt{l(l+1)} = \frac{4.716 \times 10^{-34}}{1.0545718 \times 10^{-34}} \approx 4.473. \]
03

Solve for \(l\)

We have \( \sqrt{l(l+1)} = 4.473 \). Square both sides to remove the square root: \[ l(l+1) = (4.473)^2 \approx 20.007. \] Solve the quadratic equation \( l^2 + l - 20.007 = 0 \) using the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = 1\), and \(c = -20.007\).
04

Apply the quadratic formula

Using the quadratic formula, \( l = \frac{-1 \pm \sqrt{1 + 4 \times 20.007}}{2} = \frac{-1 \pm \sqrt{81.028}}{2} \approx \frac{-1 \pm 9.005}{2} \). The valid physical solution would be \( l = \frac{-1 + 9.005}{2} \approx 4.0025 \). Since \( l \) must be an integer, we round it to the nearest whole number, which is \( l = 4 \).

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Key Concepts

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

Orbital Angular Momentum
Orbital angular momentum is a fundamental concept in quantum mechanics that describes the rotational motion of electrons around the nucleus in atoms. Unlike classical mechanics, where angular momentum can take any value, in quantum mechanics it is quantized. This means that only certain discrete values, determined by quantum numbers, are allowed. Orbital angular momentum is associated with the angular-momentum quantum number, often denoted as "l". This quantum number determines the shape of the electron cloud around the nucleus and can have integer values starting from zero. The magnitude of the orbital angular momentum is given by the expression \( L = \sqrt{l(l+1)} \hbar \), where \( \hbar \) is the reduced Planck's constant.
Understanding the quantized nature of orbital angular momentum helps explain various atomic structures and spectral lines. It is key to interpreting how electrons occupy different energy levels in an atom, providing insight into the periodic table and chemical reactivity.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of very small particles, such as electrons and photons, which follow different rules than macro-scale objects. One of its core principles is the quantization of certain physical properties. This includes energies, angular momentum, and even the orbitals of electrons within atoms.
Key elements of quantum mechanics include:
  • Wave-particle duality: Particles like electrons exhibit both wave-like and particle-like properties.
  • Quantum superposition: Particles can exist in multiple states simultaneously until measured.
  • Heisenberg's uncertainty principle: It is impossible to know both the position and momentum of a particle precisely at the same time.
These principles explain phenomena that classical physics cannot, and they are essential for technologies such as lasers, semiconductors, and quantum computers. The angular-momentum quantum number discussed in the orbital angular momentum context is a direct result of these quantum rules.
Reduced Planck's Constant
The reduced Planck's constant, often represented as \( \hbar \), is a fundamental constant in quantum mechanics. It relates to the quantization of angular momentum and energy levels. Numerically, \( \hbar = 1.0545718 \times 10^{-34} \mathrm{J} \cdot \mathrm{s} \).
The formula for orbital angular momentum \( L = \sqrt{l(l+1)} \hbar \) includes \( \hbar \) as a scaling factor, showing how fundamental it is in translating the quantum mechanical equations into measurable quantities. In many equations throughout quantum mechanics, the presence of \( \hbar \) shows up, for instance, in energy calculations and uncertainty relations.
Understanding \( \hbar \) helps in bridging the conceptual gap between observable physical quantities and their quantized descriptions in quantum theory. It acts as a conversion factor between the abstract mathematics of quantum mechanics and the physically observable phenomena.
Quadratic Formula
The quadratic formula is a mathematical tool used to find solutions to quadratic equations, which are polynomials of the form \( ax^2 + bx + c = 0 \). The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the context of quantum mechanics, the quadratic formula is essential for solving equations that describe electron behavior, like calculating the angular-momentum quantum number from given angular momentum. In this problem, by expressing the equation \( l(l+1) = (4.473)^2 \) in the form \( l^2 + l - 20.007 = 0 \), the quadratic formula provides the solutions for "l".
The formula is straightforward but powerful, often unveiling two potential solutions due to the \( \pm \) symbol. In physical applications, typically only one solution is meaningful based on the context, as in the case with the angular-momentum quantum number, which must be an integer.

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Most popular questions from this chapter

(a) Show that the total number of atomic states (including different spin states) in a shell of principal quantum number \(n\) is 2\(n^{2} .[\)Hint : The sum of the first \(N\) integers \(1+2+3+\cdots+N\) is equal to \(N(N+1) / 2 .\) (b) Which shell has 50 states?

CALC A particle is described by the normalized wave function \(\psi(x, y, z)=A x e^{-\alpha x^{2}} e^{-\beta y^{2}} e^{-\gamma y^{2}},\) where \(A, \alpha, \beta,\) and \(\gamma\) are all real, positive constants. The probability that the particle will be found in the infinitesimal volume \(d x d y d z\) centered at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is \(\left|\psi\left(x_{0}, y_{0}, z_{0}\right)\right|^{2} d x d y d z z\) (a) At what value of \(x_{0}\) is the particle most likely to be found? (b) Are there values of \(x_{0}\) for which the probability of the particle being found is zero? If so, at what \(x_{0} ?\)

A hydrogen atom in the \(n=1, m_{s}=-\frac{1}{2}\) state is placed in a magnetic field with a magnitude of 0.480 T in the \(+z\) -direction. (a) Find the magnetic interaction energy (in electron volts of the electron with the field. (b) Is there any orbital magnetic dipole moment interaction for this state? Explain. Can there be an orbital magnetic dipole moment interaction for \(n \neq 1 ?\)

The normalized radial wave function for the 2\(p\) state of the hydrogen atom is \(R_{2 p}=\left(1 / \sqrt{24 a^{5}}\right) r e^{-r / 2 a} .\) After we average over the angular variables, the radial probability function becomes \(P(r) d r=\left(R_{2 p}\right)^{2} r^{2} d r .\) At what value of \(r\) is \(P(r)\) for the 2\(p\) state a maximum? Compare your results to the radius of the \(n=2\) state in the Bohr model,

A hydrogen atom in the 5\(g\) state is placed in a magnetic field of 0.600 T that is in the \(z\) -direction. (a) Into how many levels is this state split by the interaction of the atom's orbital magnetic dipole moment with the magnetic field? (b) What is the energy separation between adjacent levels? (c) What is the energy separation between the level of lowest energy and the level of highest energy?

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