Question

Calculate the relative number of 19F nuclei in the higher and lower magnetic states at 25°C in magnetic fields of (a) 2.4 T, (b) 4.69 T, and (c) 7.05 T.

Short answer

(a) The relative number of ${}^{19}\mathrm{F}$nuclei in the higher and lower magnetic states at $25°\mathrm{C}$in the magnetic field of $2.4\mathrm{T}$is $0.9999845$.

(b) The relative number of ${}^{19}\mathrm{F}$nuclei in the higher and lower magnetic states at $25°\mathrm{C}$in the magnetic field of $4.69\mathrm{T}$is $0.9999697$.

(c) The relative number of ${}^{19}\mathrm{F}$nuclei in the higher and lower magnetic states at $25°\mathrm{C}$in the magnetic field of $7.05\mathrm{T}$is $0.9999544$.

Step-by-step solution

Step 1. Given information

The temperature is $25°\mathrm{C}$.

The magnetogyric ratio for ${}^{19}\mathrm{F}$isdata-custom-editor="chemistry" $2.5181\times {10}^8{\mathrm{T}}^{-1}{\mathrm{s}}^{-1}$.

Step 2. Convert temperature degree Celcius to kelvin

$$\begin{gathered} T\left(\mathrm{K}\right)=T(°\mathrm{C})+273 \\ =25°\mathrm{C}+273 \\ =298\mathrm{K} \end{gathered}$$

Part (a) Step 1. Given information

The given magnetic field is$2.4\mathrm{T}$.

Part (a) Step 2. Calculate the relative number of F19 nuclei at 2.4 T

The ratio of the number of the nuclei in the upper magnetic energy state to the lower energy state is as follows:

$\frac{N_j}{N_o}=\exp \left(\frac{-\gamma \mathrm{h}{B}_o}{2\mathrm{\pi k}T}\right)........\left(1\right)$

Here,

$B_o$is the magnetic field.

$\gamma$is the magnetogyric ratio.

$N_j$is the number of protons at the higher energy state.

$N_o$is the number of protons at the lower energy state.

$T$is the temperature.

data-custom-editor="chemistry" $\mathrm{h}$is the Plank’s constant equals to $6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$.

data-custom-editor="chemistry" $\mathrm{k}$is the Boltzman constant equals to data-custom-editor="chemistry" $1.38\times {10}^{-23}\mathrm{J}.{\mathrm{K}}^{-1}$.

Substitute the values in equation (1) as follows:

$$\begin{gathered} \frac{N_j}{N_o}=\exp \left(\frac{-2.5181\times {10}^8{\mathrm{T}}^{-1}{\mathrm{s}}^{-1}\times 6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\times 2.4\mathrm{T}}{2\times 3.14\times 1.38\times {10}^{-23}\mathrm{J}.{\mathrm{K}}^{-1}\times 298\mathrm{K}}\right) \\ =\exp \left(\frac{-40.068\times {10}^{-3}}{2582.59}\right) \\ =\exp \left(-0.0155\times {10}^{-3}\right) \\ =0.9999845 \end{gathered}$$

Part (b) Step 1. Given information

The given magnetic field is$4.69\mathrm{T}$.

Part (b) Step 2. Calculate the relative number of F19 nuclei at 4.69 T 

Substitute the values in equation (1) as follows:

$$\begin{gathered} \frac{N_j}{N_o}=\exp \left(\frac{-2.5181\times {10}^8{\mathrm{T}}^{-1}{\mathrm{s}}^{-1}\times 6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\times 4.69\mathrm{T}}{2\times 3.14\times 1.38\times {10}^{-23}\mathrm{J}.{\mathrm{K}}^{-1}\times 298\mathrm{K}}\right) \\ =\exp \left(\frac{-78.3\times {10}^{-3}}{2582.59}\right) \\ =\exp \left(-0.0303\times {10}^{-3}\right) \\ =0.9999697 \end{gathered}$$

Part (c) Step 1. Given information 

The given magnetic field is$7.05\mathrm{T}$.

Part (c) Step 2. Calculate the relative number of F19 nuclei at 7.05 T

Substitute the values in equation (1) as follows:

$$\begin{gathered} \frac{N_j}{N_o}=\exp \left(\frac{-2.5181\times {10}^8{\mathrm{T}}^{-1}{\mathrm{s}}^{-1}\times 6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\times 7.05\mathrm{T}}{2\times 3.14\times 1.38\times {10}^{-23}\mathrm{J}.{\mathrm{K}}^{-1}\times 298\mathrm{K}}\right) \\ =\exp \left(\frac{-117.70\times {10}^{-3}}{2582.59}\right) \\ =\exp \left(-0.04557\times {10}^{-3}\right) \\ =0.9999544 \end{gathered}$$