Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The precession frequency of the protons in a laboratory NMR spectrometer is \(15.35850 \mathrm{MHz}\). The magnetic dipole moment of the proton is \(1.410608 \cdot 10^{-26} \mathrm{~J} / \mathrm{T}\), while its spin angular momentum is \(0.5272863 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\). Calculate the magnitude of the magnetic field in which the protons are immersed.

Short Answer

Expert verified
Answer: The magnitude of the magnetic field in which the protons are immersed is approximately 0.3676 T.

Step by step solution

01

Rewrite the precession frequency in Hz

The precession frequency of the protons in the NMR spectrometer is given in MHz, so we need to convert it to Hz, by multiplying the value by \(10^6\). \(15.35850 MHz = 15.35850 \times 10^6 Hz\)
02

Find the equation connecting the precession frequency, magnetic dipole moment, and the magnetic field

The equation that relates the precession frequency \(\nu\) of the protons, the magnetic dipole moment \(\mu\), and the magnetic field \(B\) is called Larmor frequency or precession frequency and is given by, \(\nu = \frac{\mu \cdot B}{h}\), where \(h\) is the Planck's constant.
03

Rearrange the equation to solve for the magnetic field

Using the equation from step 2, rearrange it to solve for the magnetic field \(B\): \(B = \frac{\nu \cdot h}{\mu}\)
04

Identify the given values and substitute them into the equation

We are given the magnetic dipole moment \(\mu = 1.410608 \cdot 10^{-26} J/T\), the spin angular momentum \(0.5272863 \cdot 10^{-34} J\cdot s\), and the precession frequency \(\nu = 15.35850 \times 10^6 Hz\). Note that \(h = 6.626 \cdot 10^{-34} Js\), which is a constant. Now, substitute these values into the equation and solve for \(B\): \(B = \frac{(15.35850 \times 10^6 Hz)(6.626 \cdot 10^{-34} Js)}{1.410608 \cdot 10^{-26} J/T}\)
05

Calculate the magnitude of the magnetic field

Perform the calculations in the equation found in step 4: \(B = \frac{(15.35850 \times 10^6)(6.626 \cdot 10^{-34})}{1.410608 \cdot 10^{-26}} = 0.3676 \, T\) The magnitude of the magnetic field in which the protons are immersed in the laboratory NMR spectrometer is approximately \(0.3676 T\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Dipole Moment
The magnetic dipole moment is a fundamental property of particles like protons, electrons, and atomic nuclei. It represents the strength of a magnet at the atomic level and its orientation within a magnetic field. A proton's magnetic dipole moment, usually denoted by \(\mu\), is what makes nuclear magnetic resonance (NMR) possible.

When you place a proton in an external magnetic field, its magnetic dipole moment interacts with the field. This interaction leads to a mechanical torque applied to the proton, causing it to precess, much like a spinning top in a gravitational field. The level of this precession is indicative of the magnetic environment the proton experiences. For calculating purposes, we use the magnetic dipole moment in the equation \( B = \frac{u \cdot h}{\mu} \) to find out the strength of the magnetic field based on precession frequency \( u \) and Planck's constant \( h \).

In real-world applications, the magnetic dipole moment is critical because it helps scientists and engineers understand and manipulate magnetic properties at the microscopic level for imaging and spectroscopy.
Spin Angular Momentum
Spin angular momentum is an intrinsic form of angular momentum carried by elementary particles, atoms, and molecules. Unlike the angular momentum that we might be familiar with in classical mechanics, which arises from objects physically rotating about an axis, spin is a quantum property that has no classical analog.

In NMR, we are concerned with the spin of particular particles like protons, which have a spin of 1/2. While the concept of quantum spin can be abstract, its effects are very real and can be measured. When a proton with spin angular momentum is placed in a magnetic field, the spin will align with the field, causing the proton to precess at a frequency directly proportional to the magnetic field strength. The value of a proton's spin angular momentum (typically denoted by \( S \) or \( \hbar \) in calculations) is what contributes to its ability to absorb or emit electromagnetic radiation at precise frequencies within an NMR spectrometer.
Larmor Frequency
The Larmor frequency, symbolized as \( u_L \), represents the precession rate of a magnetic dipole moment in a magnetic field and is a cornerstone concept in NMR. It's crucial to realize that this frequency is not just any frequency; it's the specific frequency at which the spin system, like the protons in our case, resonates or absorbs and emits energy when in the presence of a magnetic field.

The equation \( u = \frac{\mu \cdot B}{h} \) that we used in our exercise to solve for the magnetic field strength \( B \) shows that this frequency is directly proportional to the strength of the magnetic field and the magnetic dipole moment of the particle involved. The Planck constant \( h \) acts as the proportionality factor. Thus, measuring the Larmor frequency of protons in different chemical environments allows us to infer detailed information about the molecular structure and dynamics in NMR spectroscopy, which is invaluable in fields like chemistry and medicine.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following decay modes involves a transition between states of the same nucleus? a) alpha decay b) beta decay c) gamma decay d) none of the above

The binding energy of \({ }_{2}^{3} \mathrm{He}\) is lower than that of \({ }_{1}^{3} \mathrm{H}\). Provide a plausible explanation, taking into consideration the Coulomb interaction between two protons in \({ }_{2}^{3} \mathrm{He}\).

When a target nucleus is bombarded by a beam of the appropriate particles, it is possible to produce a) a less massive nucleus, but not a more massive one. b) a more massive nucleus, but not a less massive one. c) a nucleus with a smaller charge number, but not one with a larger charge number. d) a nucleus with a larger charge number, but not one with a smaller charge number. e) a nucleus with either a larger or smaller charge number.

What is more dangerous, a radioactive material with a short half-life or a long one?

a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The Earth's oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H}\). If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He}\), how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption stayed constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free