The orbital angular momentum of an electron has a magnitude of 4.716×10−344.716\$$times10^{-34} kg-m2/s. $$What is the angular momentum quantum number ll for this electron?

Understand the relationship between the orbital angular momentum and the angular momentum quantum number. The magnitude of the orbital angular momentum (L) is given by the formula: L = √(l(l+1))ℏ, where l is the angular momentum quantum number and ℏ is the reduced Planck's constant (ℏ = 1.055 × $$10^{-34}$$ ext{ J·s}). Rearrange the formula to solve for l. Start by isolating √(l(l+1)): √(l(l+1)) = rac{L}{ℏ}. Substitute the given value of L = 4.716 × $$10^{-34}$$ ext{ kg·m}^2/ ext{s} and ℏ = 1.055 × $$10^{-34}$$ ext{ J·s} into the equation: √(l(l+1)) = rac{4.716 × $$10^{-34}$$}{1.055 × $$10^{-34}$$}. Simplify the right-hand side to find the numerical value of √(l(l+1)). Then square both sides of the equation to eliminate the square root, resulting in l(l+1) = ( ext{calculated value}$$)^2. $$Solve the quadratic equation $$l^2 + l - ( $$ext{calculated value}$$)^2 = 0 to $$find the angular momentum quantum number l. Since l must be a non-negative integer, select the appropriate solution.

Ch 41: Quantum Mechanics II: Atomic Structure

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 41: Quantum Mechanics II: Atomic Structure

Problem 11

Chapter 40, Problem 11

The orbital angular momentum of an electron has a magnitude of $4.716 \times 10^{- 34}$ kg-m²/s. What is the angular momentum quantum number $l$ for this electron?

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Understand the relationship between the orbital angular momentum and the angular momentum quantum number. The magnitude of the orbital angular momentum (L) is given by the formula:

L = √(l(l+1))ℏ

, where

l

is the angular momentum quantum number and

ℏ

is the reduced Planck's constant (

ℏ = 1.055 × 10^{-34} \(\text{ J·s}\)

Rearrange the formula to solve for

l

. Start by isolating

√(l(l+1))

√(l(l+1)) = rac{L}{ℏ}

Substitute the given value of

L = 4.716 × 10^{-34} \(\text{ kg·m}\)^2/\(\text{s}\)

and

ℏ = 1.055 × 10^{-34} \(\text{ J·s}\)

into the equation:

√(l(l+1)) = rac{4.716 × 10^{-34}}{1.055 × 10^{-34}}

Simplify the right-hand side to find the numerical value of

√(l(l+1))

. Then square both sides of the equation to eliminate the square root, resulting in

l(l+1) = (\(\text{calculated value}\))^2

Solve the quadratic equation

l^2 + l - (\(\text{calculated value}\))^2 = 0

to find the angular momentum quantum number

l

. Since

l

must be a non-negative integer, select the appropriate solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Orbital Angular Momentum

Orbital angular momentum is a measure of the rotational motion of an electron around the nucleus of an atom. It is quantized and can be expressed in terms of the angular momentum quantum number (l), where the magnitude of the angular momentum is given by the formula L = √(l(l+1))ħ, with ħ being the reduced Planck's constant.

Angular Momentum Quantum Number (l)

The angular momentum quantum number (l) determines the shape of an electron's orbital and 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), influencing the electron's energy and spatial distribution.

Planck's Constant

Planck's constant (ħ) is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It plays a crucial role in quantization, allowing the calculation of angular momentum and other properties of particles at the quantum level. Its value is approximately 1.055 x 10^-34 J·s.

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Textbook Question

Consider an electron in the $N$ shell. What is the largest orbital angular momentum this electron could have in any chosen direction? Express your answers in terms of $ℏ$ and in SI units.

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Consider an electron in the $N$ shell. What is the largest spin angular momentum this electron could have in any chosen direction? Express your answers in terms of $\hslash$ and in SI units.

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In a particular state of the hydrogen atom, the angle between the angular momentum vector $\vec{L}$ and the $z$ -axis is $u equals 26.6$ °. If this is the smallest angle for this particular value of the orbital quantum number $l$ , what is $l$ ?

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Textbook Question

A hydrogen atom in a $3 p$ state is placed in a uniform external magnetic field $\vec{B}$ . Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. What field magnitude $B$ is required to split the $3 p$ state into multiple levels with an energy difference of $2.71 \times 10^{- 5}$ eV between adjacent levels?

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Textbook Question

Consider an electron in the $N$ shell. For the electron in part (c), what is the ratio of its spin angular momentum in the $z$ -direction to its orbital angular momentum in the $z$ -direction? Note: Part (c) asked for the largest orbital angular momentum this electron could have in any chosen direction.

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Textbook Question

A hydrogen atom is in a $d$ state. In the absence of an external magnetic field, the states with different $m_{l}$ values have (approximately) the same energy. Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. Calculate the splitting (in electron volts) of the ml levels when the atom is put in a $0.800$ T magnetic field that is in the $+ z$ -direction

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The orbital angular momentum of an electron has a magnitude of 4.716×10−344.716\(\times\)10^{-34} kg-m2/s. What is the angular momentum quantum number ll for this electron?

Verified video answer for a similar problem:

Key Concepts

Orbital Angular Momentum

Angular Momentum Quantum Number (l)

Planck's Constant

The orbital angular momentum of an electron has a magnitude of $4.716 \times 10^{- 34}$ kg-m²/s. What is the angular momentum quantum number $l$ for this electron?