Skip to main content
Ch 41: Quantum Mechanics II: Atomic Structure
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 41, Problem 13

Calculate, in units of UU, the magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom for states with a principal quantum number of 22, 2020, and 200200. Compare each with the value of nhnh postulated in the Bohr model. What trend do you see?

Verified step by step guidance
1
Understand the problem: The orbital angular momentum of an electron in a hydrogen atom is quantized and given by the formula \( L = \sqrt{l(l+1)} \hbar \), where \( l \) is the orbital quantum number. For a given principal quantum number \( n \), \( l \) can take integer values from 0 to \( n-1 \). The maximum orbital angular momentum corresponds to \( l = n-1 \).
Step 1: Write the formula for the maximum orbital angular momentum. Substituting \( l = n-1 \) into the formula, we get \( L_{\text{max}} = \sqrt{(n-1)n} \hbar \).
Step 2: Express \( \hbar \) in terms of the unit \( U \). In the Bohr model, \( U = \hbar \), so \( L_{\text{max}} \) in units of \( U \) becomes \( L_{\text{max}} = \sqrt{(n-1)n} U \).
Step 3: Calculate \( L_{\text{max}} \) for \( n = 2, 20, \) and \( 200 \). For each case, substitute the value of \( n \) into the formula \( L_{\text{max}} = \sqrt{(n-1)n} U \). For example, for \( n = 2 \), \( L_{\text{max}} = \sqrt{(2-1) \cdot 2} U = \sqrt{2} U \). Repeat this for \( n = 20 \) and \( n = 200 \).
Step 4: Compare \( L_{\text{max}} \) with the Bohr model prediction. In the Bohr model, the angular momentum is postulated to be \( L = nU \). For each \( n \), compare the calculated \( L_{\text{max}} \) with \( nU \). Observe that as \( n \) increases, \( L_{\text{max}} \) approaches \( nU \), indicating that the quantum mechanical result converges to the classical Bohr model prediction for large \( n \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

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. In quantum mechanics, it is quantized and given by the formula L = mvr, where m is the mass, v is the velocity, and r is the radius of the orbit. For hydrogen, the angular momentum is also expressed in terms of the principal quantum number n, with L = nħ, where ħ is the reduced Planck's constant.
Recommended video:
Guided course
06:18
Intro to Angular Momentum

Principal Quantum Number (n)

The principal quantum number (n) indicates the energy level of an electron in an atom and determines the size of the electron's orbit. It can take positive integer values (1, 2, 3, ...), with higher values corresponding to higher energy levels and larger orbits. In the context of the hydrogen atom, the maximum orbital angular momentum is directly proportional to n, leading to the relationship L = nħ.
Recommended video:
Guided course
07:19
Moles & Avogadro's Number

Bohr Model

The Bohr model of the hydrogen atom describes the electron as moving in fixed orbits around the nucleus, with quantized energy levels. According to this model, the angular momentum of the electron is quantized and given by L = nħ, where n is the principal quantum number. This model provides a foundational understanding of atomic structure and helps predict the behavior of electrons in hydrogen-like atoms, allowing for comparisons of angular momentum across different energy levels.
Recommended video:
Guided course
07:21
Ideal Gases and the Ideal Gas Law
Related Practice
Textbook Question

A hydrogen atom undergoes a transition from a 2p2p state to the 1s1s ground state. In the absence of a magnetic field, the energy of the photon emitted is 122122 nm. The atom is then placed in a strong magnetic field in the zz-direction. Ignore spin effects; consider only the interaction of the magnetic field with the atom's orbital magnetic moment. How many different photon wavelengths are observed for the 2p1s2p\(\rightarrow\)1s transition? What are the mlm_l values for the initial and final states for the transition that leads to each photon wavelength?

1
views
Textbook Question

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?

1
views
Textbook Question

A hydrogen atom in a 3p3p state is placed in a uniform external magnetic field B\(\overrightarrow{B}\). Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. What field magnitude BB is required to split the 3p3p state into multiple levels with an energy difference of 2.71×1052.71\(\times\)10^{-5} eV between adjacent levels?

3
views
Textbook Question

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

2
views
Textbook Question

A hydrogen atom is in a dd state. In the absence of an external magnetic field, the states with different mlm_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 0.8000.800 T magnetic field that is in the +z+z-direction

1
views
Textbook Question

The orbital angular momentum of an electron has a magnitude of 4.716×10344.716\(\times\)10^{-34} kg-m2/s. What is the angular momentum quantum number ll for this electron?

1
views