/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Which of the following is not a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 17

Which of the following is not a possible value for the magnitude of the orbital angular momentum in hydrogen: (a) \(\sqrt{12} \hbar\) (b) \(\sqrt{20} \hbar ;\) (c) \(\sqrt{30} \hbar ;\) (d) \(\sqrt{40} \hbar ;\) (e) \(\sqrt{56} \hbar ?\)

Short Answer

Expert verified
The orbital angular momentum values that are not possible here (while treating \(\hbar\) as a common factor) are \(\sqrt{12}\), \(\sqrt{30}\) and \(\sqrt{40}\), implying options (a), (c) and (d) are not possible values.

Step by step solution

01

Calculate for Option (a)

Calculate if \(\sqrt{12}\) can be expressed as \(\sqrt{l(l+1)}\), where \(l\) is a non-negative integer. By doing so, it's observed that there's no such integer, hence, \(\sqrt{12} \hbar\) cannot be a possible value.
02

Calculate for Option (b)

Check if \(\sqrt{20}\) can be expressed as \(\sqrt{l(l+1)}\), where \(l\) is a non-negative integer. Here one can see that when \(l=4\), \(\sqrt{4(4+1)}\) equals to \(\sqrt{20}\). Therefore, \(\sqrt{20} \hbar\) is a possible value for the magnitude of orbital angular momentum.
03

Calculate for Option (c)

Check if \(\sqrt{30}\) can be expressed as \(\sqrt{l(l+1)}\), where \(l\) is a non-negative integer. Here, there is no such integer, so \(\sqrt{30} \hbar\) cannot be a possible value.
04

Calculate for Option (d)

Check if \(\sqrt{40}\) can be expressed as \(\sqrt{l(l+1)}\), where \(l\) is a non-negative integer. Observing, when \(l=6\), \(\sqrt{6(6+1)}\) equals to \(\sqrt{42}\), not \(\sqrt{40}\), therefore \(\sqrt{40} \hbar\) cannot be a possible value.
05

Calculate for Option (e)

Check if \(\sqrt{56}\) can be expressed as \(\sqrt{l(l+1)}\), where \(l\) is a non-negative integer. If \(l=7\), \(\sqrt{7(7+1)}\) equals to \(\sqrt{56}\). Hence, \(\sqrt{56} \hbar\) is a valid possible value for the magnitude of orbital angular momentum.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron in hydrogen is in the \(5 f\) state. What possible values, in units of \(\hbar,\) could a measurement of the orbital angular momentum component on a given axis yield?

A harmonic oscillator potential of natural frequency \(\omega\) contains eight electrons and is in its lowest-energy state. (a) What is its energy? (b) What would the lowest energy be if the electrons were replaced by spin- 1 particles of the same mass?

A harmonic oscillator potential with natural frequency \(\omega\) contains \(N\) electrons and is in its state of lowest energy. Find expressions for the energy of the highest-energy electron for (a) \(N\) even and (b) \(N\) odd.

Determine the electronic configuration of copper.

Repeat Exercise 25 for the case where you know only that the principal quantum number is \(3 ;\) that is, \(l\) might have any of its possible values.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Force

Read Explanation

Classical Mechanics

Read Explanation

Dynamics

Read Explanation

Modern Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.