Chapter 13, Problem 13.1P

Interpretation Introduction

(a)

Interpretation:

The frequency required to cause spin-flipping in an NMR spectrometer if the magnetic field at the proton is $117,400\text{gauss}$ is 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.

Answer to Problem 13.1P

The frequency required to cause spin-flipping in an NMR spectrometer if the magnetic field at the proton is $117,400\text{gauss}$ is $500.1\text{MHz}$.

Explanation of Solution

The radio frequency required to cause spin flipping is given by the equation as shown below.

${\nu }_{\text{p}}=\frac{{\gamma }_{\text{H}}}{2\pi }{\text{B}}_{\text{P}}$…(1)

Where,

• ${\nu }_{\text{p}}$ is the radio frequency.

• ${\text{B}}_{\text{P}}$ is the magnetic field at protons.

• ${\gamma }_{\text{H}}$ is the gyromagnetic ratio.

Substitute the value of ${\gamma }_{\text{H}}$ as $26753\text{radians}{\text{gauss}}^{-\text{1}}{\text{s}}^{-\text{1}}$ and value of ${\text{B}}_{\text{P}}$ as $117,400\text{gauss}$ in equation (1) is shown below.

$\begin{array}{rll}{\nu }_{\text{p}}& =& \frac{26753\text{radians}{\text{gauss}}^{-\text{1}}{\text{s}}^{-\text{1}}}{2\times 3.14\text{radians}}\times 117400\text{gauss}\\ & =& \text{500127738}\text{.8}\text{Hz}\times \frac{\text{1}\text{MHz}}{{\text{10}}^{\text{6}}\text{Hz}}\\ & =& \text{500}\text{.1}\text{MHz}\end{array}$

Conclusion

The frequency required to cause spin-flipping in an NMR spectrometer if the magnetic field at the proton is $117,400\text{gauss}$ is found to be $500.1\text{MHz}$.

Interpretation Introduction

(b)

Interpretation:

The magnetic field at the proton required to cause spin-flipping in an NMR experiment if frequency imposed on the sample is $900\text{MHz}$ is 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.

Answer to Problem 13.1P

The magnetic field at the proton required to cause spin-flipping in an NMR experiment if frequency imposed on the sample is $900\text{MHz}$ is $211,266.02\text{gauss}$.

Explanation of Solution

The radio frequency required to cause spin flipping is given by the equation as shown below.

${\nu }_{\text{p}}=\frac{{\gamma }_{\text{H}}}{2\pi }{\text{B}}_{\text{P}}$…(1)

Where,

• ${\nu }_{\text{p}}$ is the radio frequency.

• ${\text{B}}_{\text{P}}$ is the magnetic field at protons.

• ${\gamma }_{\text{H}}$ is the gyromagnetic ratio.

The magnetic field is calculated by the rearranged equation is shown below.

${\text{B}}_{\text{P}}=\frac{2\pi }{{\gamma }_{\text{H}}}{\nu }_{\text{p}}$…(2)

Substitute the value of ${\gamma }_{\text{H}}$ as $26753\text{radians}{\text{gauss}}^{-\text{1}}{\text{s}}^{-\text{1}}$ and value of ${\nu }_{\text{p}}$ as $900\text{MHz}$ in equation (2) is shown below.

$\begin{array}{rll}{\text{B}}_{\text{p}}& =& \frac{2\times 3.14\text{radians}}{26753\text{radians}{\text{gauss}}^{-\text{1}}{\text{s}}^{-\text{1}}}\times 900\times {10}^6\text{Hz}\\ & =& 211,266.02\text{gauss}\end{array}$

Conclusion

The magnetic field at the proton required to cause “spin-flipping” in an NMR experiment if frequency imposed on the sample is $900\text{MHz}$ is found to be $211,266.02\text{gauss}$.