Question

The \(^{13}\) C nucleus is a spin \(1 / 2\) particle as is a proton. However, the energy splitting for a given field strength is roughly \(1 / 4\) of that for a proton. Using a 1.45 T magnet as in Example Problem \(30.6,\) what is the ratio of populations in the excited and ground spin states for \(^{13} \mathrm{C}\) at \(298 \mathrm{K} ?\)

Short answer

The ratio of populations in the excited and ground spin states for \(^{13} \mathrm{C}\) at \(298 \mathrm{K}\) using a 1.45 T magnet can be calculated through these steps: 1. Calculate the energy difference (ΔE) between excited and ground states for 13C: ΔE_13C = (1/4) * ħ * γ_proton * B 2. Determine the Boltzmann factor: Boltzmann factor = exp(-ΔE_13C / (k * 298 )) 3. Calculate the ratio of populations in excited and ground states: N_excited / N_ground = exp(-ΔE_13C / (k * 298 )) By substituting in the values and solving, we will obtain the desired ratio of populations.

Step-by-step solution

Calculate the energy difference (ΔE) between excited and ground states for 13C.

The energy difference between two nuclear spin levels can be computed using the energy splitting formula: ΔE = ħ * γ * B where ħ is the reduced Planck's constant, γ is the gyromagnetic ratio, and B is the magnetic field strength. Since the energy splitting of 13C is 1/4 that of a proton, we can write: ΔE_13C = (1/4) * ΔE_proton Given the magnetic field strength B = 1.45 T, and the gyromagnetic ratio of the proton γ_proton = 2.67 × 10^8 T^(-1) s^(-1), we can calculate the energy difference for a proton and then for 13C. ΔE_proton = ħ * γ_proton * B ΔE_13C = (1/4) * ΔE_proton

Determine the Boltzmann factor.

The Boltzmann factor can be calculated using the formula: Boltzmann factor = exp(-ΔE / (k * T)) Where ΔE is the energy difference between the two spin states, k is Boltzmann's constant (1.38 × 10^(-23) J K^(-1)), and T is the temperature (298 K). Using the energy difference ΔE_13C calculated in Step 1, we can compute the Boltzmann factor as: Boltzmann factor = exp(-ΔE_13C / (k * 298 ))

Calculate the ratio of populations in excited and ground states.

The ratio of populations in the excited and ground states can be determined using the Boltzmann distribution: N_excited / N_ground = exp(-ΔE / (k * T)) Using the Boltzmann factor that we calculated in Step 2, we can compute the ratio of populations as: N_excited / N_ground = exp(-ΔE_13C / (k * 298 )) This will give us the ratio of populations in the excited and ground spin states for 13C at 298 K using a 1.45 T magnet. As a final result, substitute the calculated values into the equations from Steps 1-3 to find the ratio of populations in the excited and ground spin states for \(^{13} \mathrm{C}\) at \(298 \mathrm{K}\).

Key concepts

Carbon-13

Carbon-13, commonly denoted as \(^{13}\text{C}\), is an isotope of carbon that has one more neutron than the more abundant \(^{12}\text{C}\) isotope. This additional neutron results in a difference in nuclear spin properties, which is critical in the context of Nuclear Magnetic Resonance (NMR). Carbon-13 is a spin \(\frac{1}{2}\) particle, which means it has two possible spin states. This property makes it a valuable agent for NMR spectroscopy because its magnetic properties allow it to interact differently with magnetic fields when compared to elements that don't possess such spin characteristics. Apart from being used in NMR, \(^{13}\text{C}\) is also useful in various other fields, such as organic chemistry and biochemistry, for tracing carbon atoms in molecular structures. Understanding its role in NMR helps you grasp how small differences in atomic structure can lead to significant insights into molecular behavior.

Energy Levels

In the realm of NMR, energy levels refer to the difference in energy between different nuclear spin states of an atom. When exposed to a magnetic field, particles like Carbon-13 nuclei can exist in distinct spin states, typically classified as ground and excited states. The energy difference, \(\Delta E\), between these two states is calculated using the formula:\[\Delta E = \hbar \cdot \gamma \cdot B\]where

• \(\hbar\) is the reduced Planck's constant.
• \(\gamma\) is the gyromagnetic ratio of the nucleus, which is a constant specific to the type of nucleus.
• \(B\) is the magnetic field strength applied during the NMR.

For \(^{13}\text{C}\), this energy splitting is roughly \(\frac{1}{4}\) of that for a proton, leading to smaller energy transitions and requiring different considerations during analysis. By understanding these energy differences, scientists can adjust their methods to precisely study the molecular structures containing Carbon-13.

Boltzmann Distribution

The Boltzmann Distribution is a fundamental statistical concept that describes how energy states are populated among particles in a system at thermal equilibrium. In NMR, it helps determine the population ratio of nuclei in high and low energy states. The distribution can be expressed mathematically as:\[N_{\text{excited}} / N_{\text{ground}} = \exp\left(-\frac{\Delta E}{k \cdot T}\right)\]where

• \(N_{\text{excited}}\) and \(N_{\text{ground}}\) are the numbers of nuclei in the excited and ground states, respectively.
• \(\Delta E\) is the energy difference calculated from the energy levels.
• \(k\) is the Boltzmann constant, a universal constant relating energy at the particle level with temperature.
• \(T\) represents the absolute temperature of the system in Kelvin.

When applied to \(^{13}\text{C}\) in a magnetic field, this distribution formula reveals how a small fraction of nuclei populate the excited state, with most remaining in the ground state. Understanding how this distribution influences NMR signals is essential for interpreting spectra and obtaining accurate molecular information.