Question

The amount of energy required to spin-flip a nucleus depends both on the strength of the external magnetic field and on the nucleus. At a field strength of \(4.7 \mathrm{~T}, \mathrm{rf}\) energy of \(200 \mathrm{MHz}\) is required to bring a \({ }^{1} \mathrm{H}\) nucleus into resonance but energy of only \(187 \mathrm{MHz}\) will bring a \(19 \mathrm{~F}\) nucleus into resonance. Calculate the amount of energy required to spin-flip a \({ }^{19} \mathrm{~F}\) nucleus. Is this amount greater or less than that required to spin-flip a \({ }^{1} \mathrm{H}\) nucleus?

Short answer

The energy for \( ^{19} \mathrm{F} \) is less than for \( ^{1} \mathrm{H} \).

Step-by-step solution

Understand the Relationship

The energy required to spin-flip a nucleus is directly proportional to the frequency and the external magnetic field strength. The formula for the energy of a photon is given by:\[ E = h u \]where \( E \) is the energy, \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\), and \( u \) is the frequency in Hz.

Convert Frequency to Hz

Given the frequency for \( ^{19} \mathrm{F} \) is \( 187 \) MHz, convert this to Hz:\[ 187 \text{ MHz} = 187 \times 10^6 \text{ Hz} \]

Calculate Energy for Fluorine

Use the energy formula to calculate the energy required for \( ^{19} \mathrm{F} \):\[ E_{^{19}F} = h \cdot u_{^{19}F} = 6.626 \times 10^{-34} \text{ Js} \times 187 \times 10^6 \text{ Hz} \]Calculate:\[ E_{^{19}F} = 1.239 \times 10^{-25} \text{ J} \]

Compare with Hydrogen Energy

Similarly, the energy for \( ^{1} \mathrm{H} \) is calculated the same way with \( 200 \) MHz:\[ E_{^{1}H} = 6.626 \times 10^{-34} \text{ Js} \times 200 \times 10^6 \text{ Hz} = 1.325 \times 10^{-25} \text{ J} \]

Determine Greater or Less

The energy required for \( ^{19} \mathrm{F} \) is \( 1.239 \times 10^{-25} \text{ J} \), which is less than the energy required for \( ^{1} \mathrm{H} \), which is \( 1.325 \times 10^{-25} \text{ J} \).

Key concepts

Spin-flip Energy

Nuclear Magnetic Resonance (NMR) often involves a phenomenon known as 'spin-flip'. This term refers to a change in the orientation of the magnetic moment of a nucleus in the presence of an external magnetic field. The magnetic field causes the nucleus to align with it, but when enough energy is provided, this alignment flips—or "spins"—to the opposite direction. This energy, known as spin-flip energy, depends on the type of nucleus and the strength of the magnetic field involved.

In NMR, each type of nucleus will require a different amount of energy for a spin-flip to occur. This specific energy is often absorbed in the form of radio frequency (rf) waves. For instance, an NMR spectrometer might show that at a magnetic field strength of 4.7 T (tesla), a frequency of 200 MHz is needed for a hydrogen ( $^{1} ext{H}$) nucleus spin-flip. Likewise, 187 MHz is required for fluorine ($^{19} ext{F}$). Understanding specific spin-flip energies is essential for identifying and studying various atomic nuclei within substances.

Magnetic Field Strength

The magnetic field strength plays a crucial role in determining the spin-flip energy in NMR. Essentially, the stronger the magnetic field applied, the higher the frequency required to induce a spin-flip. This is because a stronger magnetic field causes a greater difference in energy between the aligned and anti-aligned states of a nucleus, according to the principles of NMR.

Here's how it works:

• An external magnetic field is applied to a sample.
• Nuclei within the sample align with or against the magnetic field.
• The energy difference becomes larger as the magnetic field strength increases.
• Increased rf frequency is required to cause a spin-flip.

Setting the right magnetic field strength is key when conducting NMR experiments. It defines the sensitivity and resolution of the spectroscopic analysis, allowing for more precise measurements of nuclear interactions and bonding environments across different chemical compounds.

Photon Energy Calculation

To calculate photon energy in NMR, we use Planck's equation, where energy ( $E$ ) is the product of Planck's constant ( $h$ ) and the frequency ( $u$ ) of the radiation. Planck's constant is a fundamental quantity in physics, valued at $6.626 imes 10^{-34} ext{ Js}$ .

In the context of NMR:

• Convert the given frequency in MHz to Hz by multiplying by $10^6$ . For example, 200 MHz becomes 200 million Hz.
• Multiply the frequency by Planck's constant to find the photon energy.
• Repeat this for each nucleus studied, such as hydrogen and fluorine.

For a practical example, calculating the energy for a $^{19}$F nucleus involves converting 187 MHz to Hz, then multiplying by Planck's constant. This yields an energy of approximately $1.239 imes 10^{-25} ext{ J}$ . Comparisons with the energy for a $^{1}$H nucleus, calculated at $1.325 imes 10^{-25} ext{ J}$ , reveal that $^{19}$F requires less energy, providing a useful context for interpreting NMR results.