Chapter 19, Problem 19.11QAP

Interpretation Introduction

(a)

Interpretation:

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 2.4 T should be calculated.

Concept introduction:

Nuclear magnetic reasonance is a technique that is used to predict the structural formula of the compound. NMR spectroscopy involves the examination of the nucleus under the external magnetic field.

Answer to Problem 19.11QAP

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 2.4 T is 0.9999845.

Explanation of Solution

The temperature and value of $B_0$, magnetic field are given as $\text{25°C}$ and 2.4 T respectively.

The conversion of temperature from Celsius into Kelvin is shown below:

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

Therefore, the temperature is $\text{298}\text{K}$.

The ratio of number of the nuclei in the upper magnetic energy state to the lower energy state is calculated by the formula,

$\frac{N_j}{N_0}=\exp \left(\frac{-\gamma h{B}_0}{2\pi kT}\right)$

Where,

• $B_0$ is the magnetic field.
• $\gamma$ is the magnetogyric ratio.
• $N_j$ is the number of protons at the higher energy state.
• $N_0$ is the number of protons at the lower energy state.
• $T$ is the temperature.
• $h$ is the Plank’s constant equals to $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ .
• $k$ is the Boltzman constant equals to $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$ .

The magnetogyric ratio for ${}^{\text{19}}\text{F}$ is taken as $2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}$ from Table 19.1.

Substitute magnetogyric ratio, Plant’s constant, Boltzmann constant, temperature and magnetic field in above expression.

$\begin{array}{c}\frac{N_j}{N_0}=\exp \left(\frac{-\left(2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}\right)\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.4\text{T}\right)}{2\left(3.14\right)\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\right)\\ =\exp \left(-\frac{40.0680072\times {10}^{-3}}{2582.5872}\right)\\ =\exp \left(-0.0155\times {10}^{-3}\right)\\ =0.9999845\end{array}$

Therefore, the relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 2.4 T is 0.9999845.

Interpretation Introduction

(b)

Interpretation:

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 4.69 T should be calculated.

Concept introduction:

Nuclear magnetic reasonance is a technique that is used to predict the structural formula of the compound. NMR spectroscopy involves the examination of the nucleus under the external magnetic field.

Answer to Problem 19.11QAP

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 4.69 T is 0.9999697.

Explanation of Solution

The temperature and value of $B_0$, magnetic field are given as $\text{25°C}$ and 4.69 T respectively.

The conversion of temperature from Celsius into Kelvin is shown below:

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

Therefore, the temperature is $\text{298}\text{K}$.

The ratio of number of the nuclei in the upper magnetic energy state to the lower energy state is calculated by the formula,

$\frac{N_j}{N_0}=\exp \left(\frac{-\gamma h{B}_0}{2\pi kT}\right)$

Where,

• $B_0$ is the magnetic field.
• $\gamma$ is the magnetogyric ratio.
• $N_j$ is the number of protons at the higher energy state.
• $N_0$ is the number of protons at the lower energy state.
• $T$ is the temperature.
• $h$ is the Plank’s constant equals to $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ .
• $k$ is the Boltzman constant equals to $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$ .

The magnetogyric ratio for ${}^{\text{19}}\text{F}$ is taken as $2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}$ from Table 19.1.

Substitute magnetogyric ratio, Plant’s constant, Boltzmann constant, temperature and magnetic field in above expression.

$\begin{array}{c}\frac{N_j}{N_0}=\exp \left(\frac{-\left(2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}\right)\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(4.69\text{T}\right)}{2\left(3.14\right)\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\right)\\ =\exp \left(-\frac{78.29956407\times {10}^{-3}}{2582.5872}\right)\\ =\exp \left(-0.0303\times {10}^{-3}\right)\\ =0.9999697\end{array}$

Therefore, the relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 4.69 T is 0.9999697.

Interpretation Introduction

(c)

Interpretation:

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 7.05 T should be calculated.

Concept introduction:

Nuclear magnetic reasonance is a technique that is used to predict the structural formula of the compound. NMR spectroscopy involves the examination of the nucleus under the external magnetic field.

Answer to Problem 19.11QAP

The relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 7.05 T is 0.9999544.

Explanation of Solution

The temperature and value of $B_0$, magnetic field are given as $\text{25°C}$ and 7.05 T respectively.

The conversion of temperature from Celsius into Kelvin is shown below:

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

Therefore, the temperature is $\text{298}\text{K}$.

The ratio of number of the nuclei in the upper magnetic energy state to the lower energy state is calculated by the formula,

$\frac{N_j}{N_0}=\exp \left(\frac{-\gamma h{B}_0}{2\pi kT}\right)$

Where,

• $B_0$ is the magnetic field.
• $\gamma$ is the magnetogyric ratio.
• $N_j$ is the number of protons at the higher energy state.
• $N_0$ is the number of protons at the lower energy state.
• $T$ is the temperature.
• $h$ is the Plank’s constant equals to $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ .
• $k$ is the Boltzman constant equals to $1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}$ .

The magnetogyric ratio for ${}^{\text{19}}\text{F}$ is taken as $2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}$ from Table 19.1.

Substitute magnetogyric ratio, Plant’s constant, Boltzmann constant, temperature and magnetic field in above expression.

$\begin{array}{c}\frac{N_j}{N_0}=\exp \left(\frac{-\left(2.5181\times {10}^8{\text{T}}^{-1}{\text{s}}^{-1}\right)\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(7.05\text{T}\right)}{2\left(3.14\right)\left(1.38\times {10}^{-23}\text{J}{\text{K}}^{-1}\right)\left(298\text{K}\right)}\right)\\ =\exp \left(-\frac{117.69977115\times {10}^{-3}}{2582.5872}\right)\\ =\exp \left(-0.04557\times {10}^{-3}\right)\\ =0.9999544\end{array}$

Therefore, the relative number of ${}^{\text{19}}\text{F}$ nuclei in the higher and lower magnetic states at $\text{25°C}$ in magnetic field of 7.05 T is 0.9999544.