Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
bartleby

Concept explainers

Question
Book Icon
Chapter 12, Problem 12A.3P

(a)

Interpretation Introduction

Interpretation:

The level of enrichment required for 15N signal to have same intensity as that from 13C with its natural abundance has to be stated.

Concept introduction:

Many nuclei and electrons have spin, due to this spin magnetic moment arises.  The energy of this magnetic moment depends on the orientation of the applied magnetic field.  In NMR spectroscopy, every nucleus has a spin.  There is an angular momentum related to the spin.  The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus.  Tetramethylsilane is taken as reference.

(a)

Expert Solution
Check Mark

Answer to Problem 12A.3P

The level of enrichment required for 15N signal to have same intensity as that from 13C with its natural abundance is 6.8%.

Explanation of Solution

The natural abundance of 13C is 1.108%.

The value of γN for 13C is 6.73×107T1 s1.

The value of γN for 15N is 2.7126×107T1 s1.

The negative sign can be neglected in the calculation of a number of nuclei.

The expression of the intensity of NMR is shown below.

    IntensityNγN2B02T

Where,

  • N is the total number of nuclei.
  • B0 is the magnetic field.
  • T is the temperature.
  • γN is the magnetogyric ratio.

The number of nuclei can be written in term of level of enrichment.

When the intensity of NMR spectra of 15N and 13C is the same, then the relationship number of total spin is given by the expression as shown below.

    N(15N)(γN(15N))2B02T=N(13C)(γN(13C))2B02TN(15N)(γN(15N))2=N(13C)(γN(13C))2N(15N)=N(13C)(γN(13C))2(γN(15N))2

Substitute the value of N(13C), (γN(13C)), and (γN(15N)) in the above expression.

  N(15N)=1.108%(6.73×107T1 s1)2(2.7126×107T1 s1)2=1.108%×4.5293×1015(7.3582×1014)=6.8%_

The level of enrichment required for 15N signal to have same intensity as that from 13C with its natural abundance is 6.8%_.

(b)

Interpretation Introduction

Interpretation:

The intensity of achievable by natural abundance of 13C has to be calculated.  The intensity of achievable by 100% of 17O has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 12A.3P

The intensity of achievable by the natural abundance of 13C is proportional to 5.018×1013×B02TT1 s1 at given temperature and magnetic field.  The intensity of achievable by 100% of 17O is proportional to 3.58644×1014×B02TT1 s1 at given temperature and magnetic field.

Explanation of Solution

The expression of the intensity of NMR is shown below.

    IntensityNγN2B02T        (1)

Where,

  • N is the total number of spin.
  • B0 is the magnetic field.
  • T is the temperature.
  • γN is the magnetogyric ratio.

The natural abundance of 13C is 1.108%.

The value of γN for 13C is 6.73×107T1 s1.

The value of γN for 17O is 1.89379×107T1 s1.

The percentage enrichment of 17O is 100%.

The number of nuclei can be written in term of level of enrichment.

Substitute the value of N and γN for 13C in the equation (1).

    Intensity(1.108%)(6.73×107T1 s1)2B02T1.108×4.5293×1015×B02100×T5.018×1013×B02TT1 s1

The intensity of achievable by natural abundance of 13C is proportional to 5.018×1013×B02TT1 s1 at given temperature and magnetic field.

Substitute the value of N and γN for 17O in the equation (1).

    Intensity(100%)(1.89379×107T1 s1)2B02T100×3.58644×1014×B02100×T3.58644×1014×B02TT1 s1

The intensity of achievable by 100% of 17O is proportional to 3.58644×1014×B02TT1 s1 at given temperature and magnetic field.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 12 Solutions

Atkins' Physical Chemistry

Ch. 12 - Prob. 12A.2BECh. 12 - Prob. 12A.3AECh. 12 - Prob. 12A.3BECh. 12 - Prob. 12A.4AECh. 12 - Prob. 12A.4BECh. 12 - Prob. 12A.5AECh. 12 - Prob. 12A.5BECh. 12 - Prob. 12A.6AECh. 12 - Prob. 12A.6BECh. 12 - Prob. 12A.7AECh. 12 - Prob. 12A.7BECh. 12 - Prob. 12A.8AECh. 12 - Prob. 12A.8BECh. 12 - Prob. 12A.9AECh. 12 - Prob. 12A.9BECh. 12 - Prob. 12A.1PCh. 12 - Prob. 12A.3PCh. 12 - Prob. 12B.1DQCh. 12 - Prob. 12B.2DQCh. 12 - Prob. 12B.3DQCh. 12 - Prob. 12B.4DQCh. 12 - Prob. 12B.5DQCh. 12 - Prob. 12B.1AECh. 12 - Prob. 12B.1BECh. 12 - Prob. 12B.2AECh. 12 - Prob. 12B.2BECh. 12 - Prob. 12B.3AECh. 12 - Prob. 12B.3BECh. 12 - Prob. 12B.4AECh. 12 - Prob. 12B.4BECh. 12 - Prob. 12B.5AECh. 12 - Prob. 12B.5BECh. 12 - Prob. 12B.6AECh. 12 - Prob. 12B.6BECh. 12 - Prob. 12B.7AECh. 12 - Prob. 12B.7BECh. 12 - Prob. 12B.8AECh. 12 - Prob. 12B.8BECh. 12 - Prob. 12B.9AECh. 12 - Prob. 12B.9BECh. 12 - Prob. 12B.10AECh. 12 - Prob. 12B.10BECh. 12 - Prob. 12B.11AECh. 12 - Prob. 12B.11BECh. 12 - Prob. 12B.12AECh. 12 - Prob. 12B.12BECh. 12 - Prob. 12B.13AECh. 12 - Prob. 12B.13BECh. 12 - Prob. 12B.14AECh. 12 - Prob. 12B.14BECh. 12 - Prob. 12B.1PCh. 12 - Prob. 12B.2PCh. 12 - Prob. 12B.3PCh. 12 - Prob. 12B.5PCh. 12 - Prob. 12B.6PCh. 12 - Prob. 12B.7PCh. 12 - Prob. 12B.8PCh. 12 - Prob. 12B.9PCh. 12 - Prob. 12C.1DQCh. 12 - Prob. 12C.2DQCh. 12 - Prob. 12C.3DQCh. 12 - Prob. 12C.4DQCh. 12 - Prob. 12C.5DQCh. 12 - Prob. 12C.1AECh. 12 - Prob. 12C.1BECh. 12 - Prob. 12C.2AECh. 12 - Prob. 12C.2BECh. 12 - Prob. 12C.3AECh. 12 - Prob. 12C.3BECh. 12 - Prob. 12C.4AECh. 12 - Prob. 12C.4BECh. 12 - Prob. 12C.5AECh. 12 - Prob. 12C.5BECh. 12 - Prob. 12C.4PCh. 12 - Prob. 12C.5PCh. 12 - Prob. 12C.6PCh. 12 - Prob. 12C.10PCh. 12 - Prob. 12D.1DQCh. 12 - Prob. 12D.2DQCh. 12 - Prob. 12D.1AECh. 12 - Prob. 12D.1BECh. 12 - Prob. 12D.2AECh. 12 - Prob. 12D.2BECh. 12 - Prob. 12D.3AECh. 12 - Prob. 12D.3BECh. 12 - Prob. 12D.4AECh. 12 - Prob. 12D.4BECh. 12 - Prob. 12D.5AECh. 12 - Prob. 12D.5BECh. 12 - Prob. 12D.6AECh. 12 - Prob. 12D.6BECh. 12 - Prob. 12D.1PCh. 12 - Prob. 12D.2PCh. 12 - Prob. 12D.4PCh. 12 - Prob. 12D.5PCh. 12 - Prob. 12D.6PCh. 12 - Prob. 12D.7PCh. 12 - Prob. 12D.8PCh. 12 - Prob. 12.3IACh. 12 - Prob. 12.4IA
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY