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Chapter 40: Problem 17

In an NMR experiment, the RF source oscillates at \(34 \mathrm{MHz}\) and magnetic resonance of the hydrogen atoms in the sample being investigated occurs when the external field \(\vec{B}_{\text {ext }}\) has magnitude 0.78 T. Assume that \(\vec{B}_{\text {int }}\) and \(\vec{B}_{\text {ext }}\) are in the same direction and take the proton magnetic moment component \(\mu_{z}\) to be \(1.41 \times 10^{-26} \mathrm{~J} / \mathrm{T}\). What is the magnitude of \(\vec{B}_{\text {int }} ?\)

Short Answer

Expert verified
The magnitude of \( \vec{B}_{\text{int}} \) is approximately \( 0.046 \mathrm{T} \).

Step by step solution

01

Understand the NMR Condition

In nuclear magnetic resonance (NMR) experiments, resonance occurs when the frequency of the radio frequency (RF) source matches the Larmor precession frequency of the nuclei. This is given by the formula \( \omega = \gamma B \), where \( \omega = 2 \pi f \) is the angular frequency, \( \gamma \) is the gyromagnetic ratio, and \( B \) is the total magnetic field experienced by the proton.
02

Calculate Angular Frequency

The given RF frequency is \( 34 \mathrm{MHz} \), which is equivalent to \( 34 \times 10^6 \mathrm{Hz} \). We can calculate the angular frequency \( \omega \) using \( \omega = 2 \pi f \), giving \( \omega = 2 \pi \times 34 \times 10^6 \).
03

Write the Equation for Total Magnetic Field

The total magnetic field \( B \) is the sum of the external magnetic field \( B_{\text{ext}} \) and the internal magnetic field \( B_{\text{int}} \). Therefore, \( B = B_{\text{ext}} + B_{\text{int}} \). Since both fields are in the same direction, they add up algebraically.
04

Use the Gyromagnetic Ratio

The gyromagnetic ratio \( \gamma \) for a proton is known: \( \gamma = \frac{2 \mu_z}{\hbar} \), where \( \hbar = 1.0545718 \times 10^{-34} \mathrm{Js} \). Insert \( \mu_z = 1.41 \times 10^{-26} \mathrm{J/T} \) to find \( \gamma \).
05

Solve for Internal Magnetic Field

Using \( \omega = \gamma (B_{\text{ext}} + B_{\text{int}}) \), calculate \( \gamma \) and substitute all known values to find \( B_{\text{int}} \). We rearrange the expression to find \( B_{\text{int}} = \frac{\omega}{\gamma} - B_{\text{ext}} \).
06

Calculate Gyromagnetic Ratio

First, calculate \( \gamma = \frac{2 \times 1.41 \times 10^{-26}}{1.0545718 \times 10^{-34}} \approx 2.675 \times 10^8 \mathrm{rad/s \cdot T} \). Using this \( \gamma \), substitute into the expression for \( B_{\text{int}} \).
07

Compute Internal Magnetic Field

Substitute \( \omega = 2 \pi \times 34 \times 10^6 \), \( \gamma \approx 2.675 \times 10^8 \mathrm{rad/s \cdot T} \), and \( B_{\text{ext}} = 0.78 \mathrm{T} \) into \( B_{\text{int}} = \frac{\omega}{\gamma} - B_{\text{ext}} \) to find \( B_{\text{int}} \approx 0.046 \mathrm{T} \).

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Key Concepts

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

Radio Frequency
Nuclear Magnetic Resonance (NMR) heavily relies on the concept of radio frequency (RF). In an NMR experiment, a sample is placed within a strong magnetic field, and radio frequency waves are applied to cause resonance within the sample's nuclei. The RF energy applied to the sample corresponds to the Larmor frequency, which is specific to the type of nucleus being studied and the magnetic field strength. This frequency is calculated using the formula:

- \( \omega = 2 \pi f \)

where \( \omega \) is the angular frequency and \( f \) is the frequency of the RF signal.

In the context of the given problem, the RF source operates at a frequency of \(34 \, \text{MHz}\). This accurate frequency is crucial as it matches the precession frequency of the protons in the sample, leading to NMR. This resonance condition allows scientists to gather valuable information about the structure of molecules in the sample.
Gyromagnetic Ratio
The gyromagnetic ratio, denoted by \( \gamma \), is an essential parameter in NMR experiments. It is defined as the ratio of the magnetic moment to the angular momentum of a particle and is unique for every type of nucleus. The formula for the gyromagnetic ratio is:

- \( \gamma = \frac{2 \mu_z}{\hbar} \)

where \( \mu_z \) represents the magnetic moment component of the particle, and \( \hbar \) is the reduced Planck鈥檚 constant.

For protons, like in our example with hydrogen atoms, this ratio is approximately \(2.675 \times 10^8 \, \text{rad/s路T}\). The gyromagnetic ratio links the frequency of the RF source and the magnetic field experienced, allowing scientists to determine whether the resonance condition is met. In mathematical terms, this is expressed as \( \omega = \gamma B \), where the total magnetic field \( B \) is the sum of internal and external magnetic fields.
Magnetic Fields
In NMR, magnetic fields play a vital role in determining the resonance condition. In our example, two critical magnetic fields are involved:
  • External Magnetic Field (\( B_{\text{ext}} \)): This is typically a consistent and strong magnetic field applied externally, which aligns the magnetic moments of the nuclei in one direction. Here, it is given as \(0.78 \, \text{T}\).
  • Internal Magnetic Field (\( B_{\text{int}} \)): This is generated due to interactions within the material itself. It affects the total magnetic field experienced by the nuclei.
Both these fields contribute to the total magnetic field \( B \), calculated as \( B = B_{\text{ext}} + B_{\text{int}} \). This total field is crucial for achieving the resonance condition in NMR. When the RF frequency matches the precession frequency determined by the total field, resonance occurs, giving insight into the molecular structure of the substance being studied.

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Most popular questions from this chapter

X rays are produced in an \(x\) -ray tube by electrons accelerated through an electric potential difference of \(50.0 \mathrm{kV}\). Let \(K_{0}\) be the kinetic energy of an electron at the end of the acceleration. The electron collides with a target nucleus (assume the nucleus remains stationary) and then has kinetic energy \(K_{1}=0.500 K_{0}\). (a) What wavelength is associated with the photon that is emitted? The electron collides with another target nucleus (assume it, too, remains stationary) and then has kinetic energy \(K_{2}=0.500 K_{1}\) (b) What wavelength is associated with the photon that is emitted?

Comet stimulated emission. When a comet approaches the Sun, the increased warmth evaporates water from the ice on the surface of the comet nucleus, producing a thin atmosphere of water vapor around the nucleus. Sunlight can then dissociate \(\mathrm{H}_{2} \mathrm{O}\) molecules in the vapor to \(\mathrm{H}\) atoms and \(\mathrm{OH}\) molecules. The sunlight can also excite the OH molecules to higher energy levels. When the comet is still relatively far from the Sun, the sunlight causes equal excitation to the \(E_{2}\) and \(E_{1}\) levels (Fig. \(40-28 a\) ). Hence, there is no population inversion between the two levels. However, as the comet approaches the Sun, the excitation to the \(E_{1}\) level decreases and population inversion occurs. The reason has to do with one of the many wavelengths - said to be Fraunhofer lines - that are missing in sunlight because, as the light travels outward through the Sun's atmosphere, those particular wavelengths are absorbed by the atmosphere. As a comet approaches the Sun, the Doppler effect due to the comet's speed relative to the Sun shifts the Fraunhofer lines in wavelength, apparently overlapping one of them with the wavelength required for excitation to the \(E_{1}\) level in \(\mathrm{OH}\) molecules. Population inversion then occurs in those molecules, and they radiate stimulated emission (Fig. \(40-28 b\) ). For example, as comet Kouhoutek approached the Sun in December 1973 and January \(1974,\) it radiated stimulated emission at about \(1666 \mathrm{MHz}\) during mid-January. (a) What was the energy difference \(E_{2}-E_{1}\) for that emission? (b) In what region of the electromagnetic spectrum was the emission?

A pulsed laser emits light at a wavelength of \(694.4 \mathrm{nm}\). The pulse duration is \(12 \mathrm{ps},\) and the energy per pulse is \(0.150 \mathrm{~J}\). (a) What is the length of the pulse? (b) How many photons are emitted in each pulse?

Seven electrons are trapped in a one-dimensional infinite potential well of width \(L .\) What multiple of \(h^{2} / 8 m L^{2}\) gives the energy of the ground state of this system? Assume that the electrons do not interact with one another, and do not neglect spin.

An electron is in a state with \(\ell=3\). (a) What multiple of \(h\) gives the magnitude of \(L ?\) (b) What multiple of \(\mu_{\mathrm{B}}\) gives the magnitude of \(\vec{\mu} ?\) (c) What is the largest possible value of \(m_{\ell},\) (d) what multiple of \(\hbar\) gives the corresponding value of \(L_{z}\), and (e) what multiple of \(\mu_{\mathrm{B}}\) gives the corresponding value of \(\mu_{\text {orb }, z} ?\) (f) What is the value of the semiclassical angle \(\theta\) between the directions of \(L_{z}\) and \(\vec{L} ?\) What is the value of angle \(\theta\) for \((\mathrm{g})\) the second largest possible value of \(m_{e}\) and \((\mathrm{h})\) the smallest (that is, most negative) possible value of \(m_{\ell} ?\)
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