Question

Recall that nuclear spin states in nuclear magnetic resonance are typically separated by energies of 2 × 10^-5 kJ mol^-1 to 2 × 10^-4 kJ mol^-1 . What are the ratios of occupation probability between a pair of such levels at thermal equilibrium and a temperature of 25°C?

Short answer

The ratio of probabilities is approximately $1$. Thus, the populations are almost equal.

Step-by-step solution

Nuclear magnetic resonance

When nuclei in a strong constant magnetic field are agitated by a weak oscillating magnetic field (in the near field), they respond by creating an electromagnetic signal with a frequency characteristic of the magnetic field at the nucleus, it is called nuclear magnetic resonance (NMR).

Given information and formula used

Nuclear spins are typically separated by energies$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{\text{kJ\hspace{0.17em}mol}}}^{\mathbf{-}\mathbf{1}}$and$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{\text{kJ\hspace{0.17em}\hspace{0.17em}mol}}}^{\mathbf{\text{-1}}}$

The ratio of probabilities is given by:

$\frac{{\mathbf{P}}_{\mathbf{i}}}{{\mathbf{P}}_{\mathbf{j}}}\mathbf{=}\mathbf{exp}\left(\frac{-\Delta E}{k_{B}T}\right)$

$\begin{array}{c}\mathbf{\Delta E}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}}\\ \mathbf{=}\mathbf{3}\mathbf{.32}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{J}\\ {\mathbf{k}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{Boltzmann}\mathbf{\text{'}}\mathbf{s}\mathbf{constant}\mathbf{)}\end{array}$

Calculation of ratio of probabilities

Calculate temperature on absolute scale.

$\begin{array}{l}\text{T}=25+273\\ =298\text{\hspace{0.17em}K}\end{array}$

Substitute the values in above equation.

$\begin{array}{c}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{j}}}=\exp \left(\frac{\text{-ΔE}}{{\text{k}}_{\text{B}}\text{T}}\right)\\ =\exp \left(\frac{-3.32\times {10}^{-26}}{1.38\times {10}^{-23}\times 298}\right)\\ =\exp (-0.807\times {10}^{-6})\\ \approx 1.0\end{array}$

Similarly, for${\text{ΔE\hspace{0.17em}=\hspace{0.17em}2×10}}^{\text{-4}}{\text{kJ\hspace{0.17em}\hspace{0.17em}mol}}^{\text{-1}}$ the same result holds, that is, $\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{j}}}\text{\hspace{0.17em}=\hspace{0.17em}1.0}$

Thus, the populations are almost equal.