Chapter 16, Problem 16.15E

Interpretation Introduction

Interpretation:

The $\Delta E$ values experienced by the ground state $\left({}^{\text{4}}{\text{F}}_{\text{3}/\text{2}}\right)$ energy levels of $\text{V}$ atom are to be calculated.

Concept Introduction:

The change in energy $\left(\Delta E\right)$ of the state depends on the strength of the magnetic field $\left(B\right)$ and the z-component quantum number $\left(M_L\right)$ in the presence of magnetic field. The expression for the change in energy $\left(\Delta E\right)$ is written below.

$\Delta E={\text{μ}}_B\cdot M_L\cdot B$

The value of $\left(M_L\right)$ can be positive, negative or zero. Hence, the change in energy $\left(\Delta E\right)$ can be positive, negative or zero. The change in energy $\left(\Delta E\right)$ of the state is equal to the amount of splitting.

Answer to Problem 16.15E

The $\Delta E$ values experienced by the ground state $\left({}^{\text{4}}{\text{F}}_{\text{3}/\text{2}}\right)$ energy levels of $\text{V}$ atom are $-3.10\times {10}^{-24}\text{J}$, $-1.03\times {10}^{-24}\text{J}$, $+1.03\times {10}^{-24}\text{J}$ and $+3.10\times {10}^{-24}\text{J}$.

Explanation of Solution

The $\Delta E$ values experienced by the ground state $\left({}^{\text{4}}{\text{F}}_{\text{3}/\text{2}}\right)$ energy levels of $\text{V}$ atom are calculated by the formula given below.

$\Delta E=g\cdot {\text{μ}}_B\cdot M_J\cdot B$…(1)

Where,

• ${\text{μ}}_B$ is Bohr magneto ($\text{9}\text{.274}\times {\text{10}}^{-\text{24}}\text{J}/\text{T}$).

• $\left(M_J\right)$ is the z-component quantum number.

• $g$ is Lande $g$ factor.

• $B$ is earth’s magnetic field.

The value of $g$ is calculated by the formula given below.

$g=1+\frac{j\left(j+1\right)+s\left(s+1\right)-l\left(l+1\right)}{2j\left(j+1\right)}$…(2)

Where,

• $j$, $s$ and $l$ are quantum numbers.

The values of $j$, $s$ and $l$ for $\left({}^{\text{4}}{\text{F}}_{\text{3}/\text{2}}\right)$ state are $3/2$, $3/2$ and $3$.

Substitute the values of $j$, $s$ and $l$ in equation (2).

$\begin{array}{rll}g& =& 1+\frac{3/2\left(3/2+1\right)+3/2\left(3/2+1\right)-3\left(3+1\right)}{2\times 3/2\left(3/2+1\right)}\\ & =& 1+\frac{15/4+15/4-12}{30/4}\\ & =& 1+\frac{30/4-12}{30/4}\\ & =& 1-\frac{18/4}{30/4}\end{array}$

Simplify the above equation.

$\begin{array}{rll}g& =& 1-\frac{3}{5}\\ & =& \frac{2}{5}\end{array}$

One gauss equals to $1.0\times {10}^{-4}\text{T}$. Therefore $5.57\times {10}^3\text{G}$ will be equal to $0.557\text{T}$.

The value of ${\text{μ}}_B$ and $B$ are ($\text{9}\text{.274}\times {\text{10}}^{-\text{24}}\text{J}/\text{T}$) and $0.557\text{T}$.

Substitute the values of ${\text{μ}}_B$, $B$ and $g$ in equation (1).

When $M_J=\left(-\frac{3}{2}\right)$,

$\begin{array}{rll}\Delta E& =& g\cdot {\text{μ}}_B\cdot M_J\cdot B\\ & =& \frac{2}{5}\times 9.274\times {10}^{-24}\text{J}/\text{T}\times \left(-\frac{3}{2}\right)\times 0.557\text{T}\\ & =& \left(-\frac{3}{5}\right)\times 5.165\times {10}^{-24}\text{J}\\ & =& -3.10\times {10}^{-24}\text{J}\end{array}$

When $M_J=\left(-\frac{1}{2}\right)$,

$\begin{array}{rll}\Delta E& =& g\cdot {\text{μ}}_B\cdot M_J\cdot B\\ & =& \frac{2}{5}\times 9.274\times {10}^{-24}\text{J}/\text{T}\times \left(-\frac{1}{2}\right)\times 0.557\text{T}\\ & =& \left(-\frac{1}{5}\right)\times 5.165\times {10}^{-24}\text{J}\\ & =& -1.03\times {10}^{-24}\text{J}\end{array}$

When $M_J=0$, $\Delta E$ will be zero.

When $M_J=\frac{3}{2}$,

$\begin{array}{rll}\Delta E& =& g\cdot {\text{μ}}_B\cdot M_J\cdot B\\ & =& \frac{2}{5}\times 9.274\times {10}^{-24}\text{J}/\text{T}\times \frac{3}{2}\times 0.557\text{T}\\ & =& \frac{3}{5}\times 5.165\times {10}^{-24}\text{J}\\ & =& 3.10\times {10}^{-24}\text{J}\end{array}$

When $M_J=\frac{1}{2}$,

$\begin{array}{rll}\Delta E& =& g\cdot {\text{μ}}_B\cdot M_J\cdot B\\ & =& \frac{2}{5}\times 9.274\times {10}^{-24}\text{J}/\text{T}\times \frac{1}{2}\times 0.557\text{T}\\ & =& \frac{1}{5}\times 5.165\times {10}^{-24}\text{J}\\ & =& 1.03\times {10}^{-24}\text{J}\end{array}$

Conclusion

The $\Delta E$ values experienced by the ground state $\left({}^{\text{4}}{\text{F}}_{\text{3}/\text{2}}\right)$ energy levels of $\text{V}$ atom are $-3.10\times {10}^{-24}\text{J}$, $-1.03\times {10}^{-24}\text{J}$, $+1.03\times {10}^{-24}\text{J}$ and $+3.10\times {10}^{-24}\text{J}$.