Chapter 12, Problem 12A.7AE

(i)

Interpretation Introduction

Interpretation:

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{0}\text{.30}\text{T}$ 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.

  $v=\frac{{\gamma }_{\text{N}}{B}_o}{2\text{π}}$

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.

Answer to Problem 12A.7AE

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{0}\text{.30}\text{T}$ is $\underset{\_}{1.02\times 10^{-6}}$.

Explanation of Solution

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{N_{\alpha }-N_{\beta }}{N}\approx \frac{{\gamma }_{\text{N}}\hslash B_0}{2kT}\\ =\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\end{array}$        (1)

Where,

• ${\mu }_{\text{N}}$ is the nuclear magneton.
• $g_I$ is the nuclear $g-$ factor.
• $B_o$ is the magnetic field.
• $N$ is the total number of spin.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The temperature is $25°\text{C}$.  The conversion of temperature in Kelvin is shown below.

  $\begin{array}{c}T=\left(273+25°\text{C}\right)\text{K}\\ =298\text{ K}\end{array}$

The value if ${\mu }_{\text{N}}$ is $5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}$, the value of $g_I$ is $5.5857$, the magnetic field $\left(B_o\right)$ is $\text{0}\text{.30}\text{T}$, the value of $k$ is $1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}$ and the temperature is $298\text{ K}$.

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\\ =\frac{\left(5.5857\right)\left(5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}\right)\left(0.30\text{T}\right)}{2\left(1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\\ =\frac{8.46\times {10}^{-27}}{8.23\times {10}^{-21}}\\ =\underset{\_}{1.02\times 10^{-6}}\end{array}$

Therefore, the relative population differences for the ${}^{\text{1}}\text{H}$ nuclei is $\underset{\_}{1.02\times 10^{-6}}$.

(ii)

Interpretation Introduction

Interpretation:

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{1}\text{.5}\text{T}$ 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.

  $v=\frac{{\gamma }_{\text{N}}{B}_o}{2\text{π}}$

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.

Answer to Problem 12A.7AE

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{1}\text{.5}\text{T}$ is $\underset{\_}{5.14\times 10^{-6}}$.

Explanation of Solution

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{N_{\alpha }-N_{\beta }}{N}\approx \frac{{\gamma }_{\text{N}}\hslash B_0}{2kT}\\ =\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\end{array}$        (1)

Where,

• ${\mu }_{\text{N}}$ is the nuclear magneton.
• $g_I$ is the nuclear $g-$ factor.
• $B_o$ is the magnetic field.
• $N$ is the total number of spin.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The temperature is $25°\text{C}$.  The conversion of temperature in Kelvin is shown below.

  $\begin{array}{c}T=\left(273+25°\text{C}\right)\text{K}\\ =298\text{ K}\end{array}$

The value if ${\mu }_{\text{N}}$ is $5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}$, the value of $g_I$ is $5.5857$, the magnetic field $\left(B_o\right)$ is $\text{1}\text{.5}\text{T}$, the value of $k$ is $1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}$ and the temperature is $298\text{ K}$.

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\\ =\frac{\left(5.5857\right)\left(5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}\right)\left(\text{1}\text{.5}\text{T}\right)}{2\left(1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\\ =\frac{42.32\times {10}^{-27}}{8.23\times {10}^{-21}}\\ =\underset{\_}{5.14\times 10^{-6}}\end{array}$

Therefore, the relative population differences for the ${}^{\text{1}}\text{H}$ nuclei is $\underset{\_}{5.14\times 10^{-6}}$.

(iii)

Interpretation Introduction

Interpretation:

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{10}\text{T}$ 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.

  $v=\frac{{\gamma }_{\text{N}}{B}_o}{2\text{π}}$

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.

Answer to Problem 12A.7AE

The relative population differences for the ${}^{\text{1}}\text{H}$ nuclei in the magnetic field of $\text{10}\text{T}$ is $\underset{\_}{3.42\times 10^{-5}}$.

Explanation of Solution

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{N_{\alpha }-N_{\beta }}{N}\approx \frac{{\gamma }_{\text{N}}\hslash B_0}{2kT}\\ =\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\end{array}$        (1)

Where,

• ${\mu }_{\text{N}}$ is the nuclear magneton.
• $g_I$ is the nuclear $g-$ factor.
• $B_o$ is the magnetic field.
• $N$ is the total number of spin.
• $k$ is the Boltzmann constant.
• $T$ is the temperature.

The temperature is $25°\text{C}$.  The conversion of temperature in Kelvin is shown below.

  $\begin{array}{c}T=\left(273+25°\text{C}\right)\text{K}\\ =298\text{ K}\end{array}$

The value if ${\mu }_{\text{N}}$ is $5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}$, the value of $g_I$ is $5.5857$, the magnetic field $\left(B_o\right)$ is $\text{10}\text{T}$, the value of $k$ is $1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}$ and the temperature is $298\text{ K}$.

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

  $\begin{array}{c}\frac{\delta N}{N}=\frac{g_{\text{I}}{\mu }_{\text{N}}{B}_0}{2kT}\\ =\frac{\left(5.5857\right)\left(5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}\right)\left(\text{10}\text{T}\right)}{2\left(1.381\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\\ =\frac{28.21\times {10}^{-26}}{8.23\times {10}^{-21}}\\ =\underset{\_}{3.42\times 10^{-5}}\end{array}$

Therefore, the relative population differences for the ${}^{\text{1}}\text{H}$ nuclei is $\underset{\_}{3.42\times 10^{-5}}$.