Chapter 12, Problem 12A.4AE

Interpretation Introduction

Interpretation:

The energies of the nuclear spin state of $\text{S}$ in magnetic field of $\text{6}\text{.8}\text{T}$ have to be stated.

Concept introduction:

The frequency of electromagnetic radiation is given by resonance condition.  This resonance condition gives an equation that relates the frequency of nuclei with the magnetic field.  The frequency of a nuclei is given by the equation as shown below.

  $v=\frac{{\gamma }_{\text{N}}{B}_o}{2\text{π}}$

The above stated frequency is known as the Larmor precession frequency.  Larmor precession is a phenomenon in which the magnetic moment of nuclei precesses about an external magnetic field.

Answer to Problem 12A.4AE

The energies of the nuclear spin state of $\text{S}$ in magnetic field of $\text{6}\text{.8}\text{T}$ are $\underset{\_}{-2.21\times 10^{-26}J}$, $\underset{\_}{-7.365\times 10^{-27}J}$, $\underset{\_}{7.365\times 10^{-27}J}$, and $\underset{\_}{2.21\times 10^{-26}J}$.

Explanation of Solution

The energies of the nuclear spin in a magnetic field is calculated by the formula shown below.

    $E_{m_I}=-g_I{\mu }_{\text{N}}{B}_0{m}_I$        (1)

Where,

• ${\mu }_{\text{N}}$ is the nuclear magneton $\left(5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}\right)$.
• $g_I$ is the nuclear $g-$ factor.
• $m_I$ is the possible value of spin quantum number.
• $B_o$ is the magnetic field.

For $\text{S}$, the spin quantum number $\left(I\right)$ is $\frac{3}{2}$.  Therefore, the possible values of spin quantum number $\left(m_I\right)$ for $\text{S}$ are calculated as shown below.

    $m_I=I,\left(I-1\right),\left(I-2\right),\left(I-3\right)$        (2)

Substitute the value of the spin quantum number $\left(I\right)$ as $\frac{3}{2}$ in equation (2).

    $\begin{array}{c}{m}_I=I,\left(I-1\right),\left(I-2\right),\left(I-3\right)\\ =\frac{3}{2},\left(\frac{3}{2}-1\right),\left(\frac{3}{2}-2\right),\left(\frac{3}{2}-3\right)\\ =\frac{3}{2},\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\end{array}$

The value of nuclear $g-$ factor and magnetic field $\left(B_o\right)$ is $0.4289$ and $\text{6}\text{.8}\text{T}$ respectively.

Substitute the value of nuclear $g-$ factor, magnetic field $\left(B_o\right)$, and nuclear magneton $\left({\mu }_{\text{N}}\right)$ in equation (1).

    $\begin{array}{c}{E}_{m_I}=-g_I{\mu }_{\text{N}}{B}_0{m}_I\\ =-0.4289\times \left(5.051\times {10}^{-27}\text{J}{\text{T}}^{-1}\right)\times \text{6}\text{.8}\text{T}\times m_I\\ =-m_I\left(14.73\times {10}^{-27}\text{J}\right)\\ =-m_I\left(1.473\times {10}^{-26}\text{J}\right)\end{array}$

At $m_I=\frac{3}{2}$, the energy of the nuclear spin in a magnetic field is calculated as shown below.

    $\begin{array}{c}{E}_{m_I}=-m_I\left(1.473\times {10}^{-26}\text{J}\right)\\ =-\frac{3}{2}\times 1.473\times {10}^{-26}\text{J}\\ =-\underset{\_}{2.21\times 10^{-26}J}\end{array}$

At $m_I=\frac{1}{2}$, the energy of the nuclear spin in a magnetic field is calculated as shown below

    $\begin{array}{c}{E}_{m_I}=-m_I\left(1.473\times {10}^{-26}\text{J}\right)\\ =-\frac{1}{2}\times 1.473\times {10}^{-26}\text{J}\\ =-0.7365\times {10}^{-26}\text{J}\\ =\underset{\_}{-7.365\times 10^{-27}J}\end{array}$

At $m_I=-\frac{1}{2}$, the energy of the nuclear spin in a magnetic field is calculated as shown below

    $\begin{array}{c}{E}_{m_I}=-m_I\left(1.473\times {10}^{-26}\text{J}\right)\\ =-\left(-\frac{1}{2}\right)\times 1.473\times {10}^{-26}\text{J}\\ =0.7365\times {10}^{-26}\text{J}\\ =\underset{\_}{7.365\times 10^{-27}J}\end{array}$

At $m_I=-\frac{3}{2}$, the energy of the nuclear spin in a magnetic field is calculated as shown below.

    $\begin{array}{c}{E}_{m_I}=-m_I\left(1.473\times {10}^{-26}\text{J}\right)\\ =-\left(-\frac{3}{2}\right)\times 1.473\times {10}^{-26}\text{J}\\ =\underset{\_}{2.21\times 10^{-26}J}\end{array}$