Question

Complete the following table. Which nucleus requires the least energy to flip its spin at this applied field? Which nucleus requires the most energy? $$ \begin{array}{|c|c|c|c|} \hline \text { Nucleus } & \begin{array}{c} \text { Applied Field } \\ \text { (tesla, T) } \end{array} & \begin{array}{c} \text { Radio Frequency } \\ (\mathbf{M H z}) \end{array} & \text { Energy (J/mol) } \\ \hline{ }^{1} \mathrm{H} & 7.05 & 300 & \\ \hline{ }^{13} \mathrm{C} & 7.05 & 75.5 & \\ \hline{ }^{19} \mathrm{~F} & 7.05 & 282 & \\ \hline \end{array} $$

Short answer

The nucleus that requires the least energy to flip the spin is the \(^{13}C\) nucleus, with only 14.80 J/mol. The nucleus that requires the most energy to flip the spin is the \(^{1}H\) nucleus, needing 59.38 J/mol.

Step-by-step solution

Convert radio frequency to Hz

For each nucleus, multiply the given radio frequency value by \(10^6\) to obtain the frequency in Hz: - \(1H\): \(300 \times 10^6 \, Hz\) - \(^{13}C\): \(75.5 \times 10^6 \, Hz\) - \(^{19}F\): \(282 \times 10^6 \, Hz\)

Calculate the gyromagnetic ratio

Using the frequency (Hz) and the magnetic field (T) from the table, calculate the gyromagnetic ratio of the nucleus using the formula \(\gamma = 2\pi \nu / B\): - \(1H\): \(\gamma_{1H} = 2\times\pi\times300\times10^6 / 7.05 =267.515\times10^6 \, \mathrm{Hz/T}\) - \(^{13}C\): \(\gamma_{13C} = 2\times\pi\times75.5\times10^6 / 7.05 = 67.172\times10^6 \, \mathrm{Hz/T}\) - \(^{19}F\): \(\gamma_{19F} = 2\times\pi\times282\times10^6 / 7.05 = 251.867\times10^6 \, \mathrm{Hz/T}\)

Calculate the energy

Now, using the gyromagnetic ratio and the magnetic field, calculate the energy for each nucleus using the formula \(E = 2 \cdot \pi \cdot h \cdot (\gamma / 2 \pi) B\): - \(1H\): \(E_{1H} = 2 \cdot \pi \cdot 6.626\times10^{-34} \cdot (267.515\times10^6) \cdot 7.05 = 9.861\times10^{-25} \mathrm{J}\) - \(^{13}C\): \(E_{13C} = 2 \cdot \pi \cdot 6.626\times10^{-34} \cdot (67.172\times10^6) \cdot 7.05 = 2.459\times10^{-25} \mathrm{J}\) - \(^{19}F\): \(E_{19F} = 2 \cdot \pi \cdot 6.626\times10^{-34} \cdot (251.867\times10^6) \cdot 7.05 = 9.306\times10^{-25} \mathrm{J}\) Now, converting energy to J/mol by multiplying with Avogadro's number (\(6.022\times10^{23} \mathrm{mol^{-1}}\)): - \(1H\): \(E_{1H} = 9.861\times10^{-25} \mathrm{J}\cdot6.022\times10^{23}\mathrm{mol^{-1}}= 59.38 \mathrm{J/mol}\) - \(^{13}C\): \(E_{13C} = 2.459\times10^{-25} \mathrm{J}\cdot6.022\times10^{23}\mathrm{mol^{-1}}= 14.80 \mathrm{J/mol}\) - \(^{19}F\): \(E_{19F} = 9.306\times10^{-25} \mathrm{J}\cdot6.022\times10^{23}\mathrm{mol^{-1}}= 56.01 \mathrm{J/mol}\) The least energy is required for \(^{13}C\) nucleus (14.80 J/mol), and the most energy is required for the \(^{1}H\) nucleus (59.38 J/mol).

Key concepts

Spin Flip Energy Calculation

The concept of spin flip energy calculation in Nuclear Magnetic Resonance (NMR) spectroscopy is pivotal to understanding how different nuclei respond to a magnetic field. When a nucleus with a magnetic moment is placed in a magnetic field, it can align either with or against the field. These two states have different energy levels, and the 'spin flip' refers to the transition of the nucleus from the lower to the higher energy state.

To determine the energy required for this transition, we use the formula: \[\begin{equation} E = h u \end{equation}\], where \(E\) is the energy in joules (J), \(h\) is Planck's constant \(6.626 \times 10^{-34} J \cdot s\), and \(u\) is the frequency of the radio waves in hertz (Hz) that match the energy difference between the two states. In NMR spectroscopy, the energy required for a spin flip (transition) is provided by radiofrequency (RF) radiation. The exact frequency needed is unique for each type of nucleus and depends on the magnetic field strength.

During a spin flip, the nucleus absorbs a quantum of energy precisely matching this frequency, known as the resonant frequency. The energy absorption is subject to quantization—only specific amounts of energy corresponding to the nuclear spins' transition can be absorbed, which is given by the above equation. This property allows NMR spectroscopy to identify different elements within a sample and to determine molecular structures and dynamics.

Gyromagnetic Ratio

The gyromagnetic ratio \(\gamma\) is a fundamental property of atomic nuclei with magnetic moment. It relates to how the nucleus interacts with an external magnetic field and is defined as the ratio of the magnetic moment \(\mu\) to the angular momentum \(L\). In a more practical sense for NMR, it connects the frequency of the applied radio frequency (RF) radiation to the magnetic field strength at which resonance occurs.

To calculate the gyromagnetic ratio, the formula \[\begin{equation} \gamma = \frac{2\piu}{B} \end{equation}\] is used, where \(u\) is the resonant frequency in Hz and \(B\) is the strength of the magnetic field in tesla (T). This relationship is paramount in NMR, as it dictates which RF frequency will be required to flip a nucleus's spin when a specific magnetic field is applied. It's notable that each type of nucleus has a unique gyromagnetic ratio, meaning that the energy needed to flip its spin at a given magnetic field will also be unique.

The gyromagnetic ratio is an intrinsic property of a nucleus and does not change, regardless of the environment. Knowledge of the gyromagnetic ratio allows chemists and physicists to set the correct frequency for the NMR instrument to achieve resonance for the specific nuclei being examined.

Radio Frequency in NMR

In Nuclear Magnetic Resonance (NMR) spectroscopy, radio frequency (RF) plays a central role. It is the energy form used to induce transitions (or 'spin flips') between nuclear energy levels. When these transitions occur, nuclei absorb or release energy, which can then be detected and measured.

The RF required for these transitions is specific to the nucleus being observed and is determined by both the gyromagnetic ratio and the strength of the magnetic field according to the Larmor equation: \[\begin{equation} u = \frac{\gamma B}{2\pi} \end{equation}\]
For instance, if the gyromagnetic ratio of hydrogen is known, and the magnetic field strength is set, we can calculate the exact frequency of the RF radiation required to flip the spin of hydrogen nuclei. This calculated frequency is what is used to perturb the nuclei during an NMR experiment. If the frequency matches the energy difference between the nuclear magnetic energy levels, absorption occurs, which is detected by the NMR spectrometer.

Understanding the importance of the precise RF frequency allows students to grasp why careful calibration and adjustment of the NMR instrument are essential for accurate spectroscopic measurements. The frequency must be tuned specifically for the magnetic field strength and the particular nucleus being observed to ensure successful spin flips and reliable data.