Chapter 12, Problem 12C.4P

(a)

Interpretation Introduction

Interpretation:

The width at half height for a Lorentzian function is $\frac{1}{\text{π}{T}_2}$ has to be derived.

Concept introduction:

In Fourier transform NMR, the nuclear spins are exposed to radiofrequency radiation.  This radiation has frequency near to or equal to the Larmor frequency which leads to excitation of spins.  The spins on reaching back to equilibrium emit radiation.  This emitted signal is recorded with respect to time and the signal is fourier transformed and computed. The transfer of energy can occur within the spin and the lattic, this process is known as spin-lattice relaxation.  If the energy of magnetization is exchanged within the nuclear spins it ics known as spin-spin relaxations.

Answer to Problem 12C.4P

The width at half height for a Lorentzian function is $\frac{1}{\text{π}{T}_2}$ has been derived.

Explanation of Solution

The Lorentzian function is given by the equation.

  $I_L\left(\omega \right)=\frac{S_o{T}_2}{1+T_2^2{\left(\omega -{\omega }_o\right)}^2}$        (1)

Where,

• $S_o$ is a constant.
• $T_2$ is spin-spin relaxation time.
• $\omega$ is the angular frequency.
• ${\omega }_o$ is the resonance frequency.

The change in the frequency can be given by $\Delta \omega$, therefore equation (1) is written as shown below.

  $I_L\left(\omega \right)=\frac{S_o{T}_2}{1+T_2^2{\left(\Delta \omega \right)}^2}$        (2)

At maxima, $\omega ={\omega }_o$ and equation (1) reduces to the equation as shown below.

  $I_{L,\max }=S_o{T}_2$        (3)

At half height, the intensity is given by the equation shown below.

  $I_L\left(\Delta {\omega }_{1/2}\right)=\frac{I_{L,\max }}{2}$        (4)

Substitute equation (3) in equation (4).

  $I_L\left(\Delta {\omega }_{1/2}\right)=\frac{S_o{T}_2}{2}$        (5)

The intensity at half height is also given by the expression.

  $I_L\left(\Delta {\omega }_{1/2}\right)=\frac{S_o{T}_2}{1+T_2^2{\left({\omega }_{1/2}-\omega \right)}^2}$        (6)

Equate equation (2) and (5) to evaluate the change is frequency.

  $\begin{array}{c}\frac{S_o{T}_2}{1+T_2^2{\left(\Delta \omega \right)}^2}=\frac{1}{2}\\ 1+T_2^2{\left(\Delta \omega \right)}^2=2\end{array}$

The above equation further reduces to the equation as shown below.

  $\Delta \omega =\frac{1}{T_2}$        (7)

The relation between change in frequency $\Delta \omega$ and change is frequency $\Delta {\omega }_{1/2}$ at half height is shown below.

  $\Delta \omega =\frac{1}{2}\Delta {\omega }_{1/2}$        (8)

Equate equation (7) and (8) as shown below.

  $\begin{array}{c}\frac{1}{T_2}=\frac{1}{2}\Delta {\omega }_{1/2}\\ \Delta {\omega }_{1/2}=\frac{2}{T_2}\end{array}$

The change in angular frequency is given by the equation $\Delta \omega =2\text{π}\left(\Delta v\right)$. Substitute this in the above expression.

  $\begin{array}{c}\Delta v_{1/2}=\frac{2}{2\text{π}{T}_2}\\ =\frac{1}{\text{π}{T}_2}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The width at half height for a Gaussian function is $\frac{2{\left(\ln 2\right)}^{1/2}}{T_2}$ has to be derived.

Concept introduction:

The same concept introduction as in subpart (a).

Answer to Problem 12C.4P

The width at half height for a Gaussian function is $\frac{2{\left(\ln 2\right)}^{1/2}}{T_2}$ has been derived.

Explanation of Solution

The Gaussian function is given by the equation.

  $I_G\left(\omega \right)=S_o{T}_2{\text{e}}^{-T_2^2{\left(\omega -{\omega }_o\right)}^2}$        (9)

Where,

• $S_o$ is a constant.
• $T_2$ is spin-spin relaxation time.
• $\omega$ is the angular frequency.
• ${\omega }_o$ is the resonance frequency.

The change in the frequency can be given by $\Delta \omega$, therefore equation (1) is written as shown below.

  $I_G\left(\omega \right)=S_o{T}_2{\text{e}}^{-T_2^2{\left(\Delta \omega \right)}^2}$        (10)

At maxima, $\omega ={\omega }_o$ and equation (9) reduces to the equation as shown below.

  $I_{G,\max }=S_o{T}_2$        (11)

At half height, the intensity is given by the equation shown below.

  $I_G\left(\Delta {\omega }_{1/2}\right)=\frac{I_{G,\max }}{2}$        (12)

Substitute equation (11) in equation (12).

  $I_G\left(\Delta {\omega }_{1/2}\right)=\frac{S_o{T}_2}{2}$        (13)

The intensity at half height is also given by the expression.

  $I_G\left(\Delta {\omega }_{1/2}\right)=S_o{T}_2{\text{e}}^{-T_2^2{\left({\omega }_{1/2}-\omega \right)}^2}$        (14)

Equate equation (10) and (13) to evaluate the change is frequency.

  $\begin{array}{c}{\text{e}}^{-T_2^2{\left(\Delta \omega \right)}^2}=\frac{1}{2}\\ \ln 2=2T_2^2{\left(\Delta \omega \right)}^2\end{array}$

The above equation further reduces to the equation as shown below.

  $\left(\Delta \omega \right)=\frac{{\left(\ln 2\right)}^{1/2}}{T_2}$        (15)

The relation between change in frequency $\Delta \omega$ and change is frequency $\Delta {\omega }_{1/2}$ at half height is shown below.

  $\Delta \omega =\frac{1}{2}\Delta {\omega }_{1/2}$        (16)

Equate equation (15) and (16) as shown below.

  $\begin{array}{c}\frac{{\left(\ln 2\right)}^{1/2}}{T_2}=\frac{1}{2}\Delta {\omega }_{1/2}\\ \Delta {\omega }_{1/2}=\frac{2{\left(\ln 2\right)}^{1/2}}{T_2}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The comparison between Lorentzian and Gaussian curves has to be stated.

Concept introduction:

The same concept introduction as in subpart (i).

Answer to Problem 12C.4P

The curves for Lorentzian and Gaussian functions are shown below.

Explanation of Solution

Consider all the parameters in equation (1) as constant. Therefore, the Lorentzian function reduces to the equation as shown below.

  $I_L\left(\omega \right)=\frac{1}{1+x^2}$        (17)

Consider all the parameters in equation (9) as constant. Therefore, the Gaussian function reduces to the equation as shown below.

  $I_G\left(\omega \right)={\text{e}}^{-x^2}$        (18)

The curves for the equation (17) and (18) are shown below.

Figure 1

In the above graph, the series 1 is for Lorentzian function and series 2 is for Gaussian function.