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
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Chapter 12, Problem 12A.7AE

(i)

Interpretation Introduction

Interpretation:

The relative population differences for the 1H nuclei in the magnetic field of 0.30T has to be calculated.

Concept introduction:

The frequency of electromagnetic radiation is given by resonance condition.  This resonance condition gives an equation that relates the frequency of nuclei with the magnetic field.  The frequency of nuclei is given by the equation as shown below.

  v=γNBo2π

The above-stated frequency is known as the Larmor precession frequency.  Larmor precession is a phenomenon in which the magnetic moment of nuclei precesses about an external magnetic field.

(i)

Expert Solution
Check Mark

Answer to Problem 12A.7AE

The relative population differences for the 1H nuclei in the magnetic field of 0.30T is 1.02×106_.

Explanation of Solution

The relative population difference of the nuclei is calculated by the formula shown below.

  δNN=NαNβNγNB02kT=gIμNB02kT        (1)

Where,

  • μN is the nuclear magneton.
  • gI is the nuclear g factor.
  • Bo is the magnetic field.
  • N is the total number of spin.
  • k is the Boltzmann constant.
  • T is the temperature.

The temperature is 25 °C.  The conversion of temperature in Kelvin is shown below.

  T=(273+25 °C)K=298 K

The value if μN is 5.051×1027JT1, the value of gI is 5.5857, the magnetic field (Bo) is 0.30T, the value of k is 1.381×1023JK1 and the temperature is 298 K.

Substitute the corresponding values in equation (1) as shown below.

  δNN=gIμNB02kT=(5.5857)(5.051×1027JT1)(0.30T)2(1.381×1023JK1)(298K)=8.46×10278.23×1021=1.02×106_

Therefore, the relative population differences for the 1H nuclei is 1.02×106_.

(ii)

Interpretation Introduction

Interpretation:

The relative population differences for the 1H nuclei in the magnetic field of 1.5T has to be calculated.

Concept introduction:

The frequency of electromagnetic radiation is given by resonance condition.  This resonance condition gives an equation that relates the frequency of nuclei with the magnetic field.  The frequency of nuclei is given by the equation as shown below.

  v=γNBo2π

The above-stated frequency is known as the Larmor precession frequency.  Larmor precession is a phenomenon in which the magnetic moment of nuclei precesses about an external magnetic field.

(ii)

Expert Solution
Check Mark

Answer to Problem 12A.7AE

The relative population differences for the 1H nuclei in the magnetic field of 1.5T is 5.14×106_.

Explanation of Solution

The relative population difference of the nuclei is calculated by the formula shown below.

  δNN=NαNβNγNB02kT=gIμNB02kT        (1)

Where,

  • μN is the nuclear magneton.
  • gI is the nuclear g factor.
  • Bo is the magnetic field.
  • N is the total number of spin.
  • k is the Boltzmann constant.
  • T is the temperature.

The temperature is 25 °C.  The conversion of temperature in Kelvin is shown below.

  T=(273+25 °C)K=298 K

The value if μN is 5.051×1027JT1, the value of gI is 5.5857, the magnetic field (Bo) is 1.5T, the value of k is 1.381×1023JK1 and the temperature is 298 K.

Substitute the corresponding values in equation (1) as shown below.

  δNN=gIμNB02kT=(5.5857)(5.051×1027JT1)(1.5T)2(1.381×1023JK1)(298K)=42.32×10278.23×1021=5.14×106_

Therefore, the relative population differences for the 1H nuclei is 5.14×106_.

(iii)

Interpretation Introduction

Interpretation:

The relative population differences for the 1H nuclei in the magnetic field of 10T has to be calculated.

Concept introduction:

The frequency of electromagnetic radiation is given by resonance condition.  This resonance condition gives an equation that relates the frequency of nuclei with the magnetic field.  The frequency of nuclei is given by the equation as shown below.

  v=γNBo2π

The above-stated frequency is known as the Larmor precession frequency.  Larmor precession is a phenomenon in which the magnetic moment of nuclei precesses about an external magnetic field.

(iii)

Expert Solution
Check Mark

Answer to Problem 12A.7AE

The relative population differences for the 1H nuclei in the magnetic field of 10T is 3.42×105_.

Explanation of Solution

The relative population difference of the nuclei is calculated by the formula shown below.

  δNN=NαNβNγNB02kT=gIμNB02kT        (1)

Where,

  • μN is the nuclear magneton.
  • gI is the nuclear g factor.
  • Bo is the magnetic field.
  • N is the total number of spin.
  • k is the Boltzmann constant.
  • T is the temperature.

The temperature is 25 °C.  The conversion of temperature in Kelvin is shown below.

  T=(273+25 °C)K=298 K

The value if μN is 5.051×1027JT1, the value of gI is 5.5857, the magnetic field (Bo) is 10T, the value of k is 1.381×1023JK1 and the temperature is 298 K.

Substitute the corresponding values in equation (1) as shown below.

  δNN=gIμNB02kT=(5.5857)(5.051×1027JT1)(10T)2(1.381×1023JK1)(298K)=28.21×10268.23×1021=3.42×105_

Therefore, the relative population differences for the 1H nuclei is 3.42×105_.

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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
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