Chapter 19, Problem 18E

(a)

To determine

The energy supplied to reverse the direction of a magnetic dipole moment due to the spin of an electron.

Answer to Problem 18E

The energy required to reverse the direction of the magnetic dipole moment of an electron due to its spin is $\underset{\_}{1.15\times {10}^{-3}\text{ eV}}$.

Explanation of Solution

A loop wire carrying a current in a magnetic field is called a magnetic dipole.

Write the expression to find the potential energy of a magnetic dipole.

    $U=-\mu B\cos \theta$        (I)

Here, $U$ is the potential energy, $\mu$ is the magnetic dipole moment, $B$ is the magnitude of the field strength, and $\theta$ is the angle between the plane of the loop and the field.

The $z$ component of the magnetic dipole moment due to the spin of an electron about its axis is,

    ${\mu }_z=\pm \frac{eh}{4\pi m}$        (II)

Here, ${\mu }_z$ is the $z$ component of the magnetic dipole moment due to the spin of the electron, $e$ is the charge of an electron, $h$ is the Planck’s constant, and $m$ is the mass of the electron.

Conclusion:

Substitute $1.6\times {10}^{-19}\text{ C}$ for $e$, $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$, and $9.1\times {10}^{-31}\text{ kg}$ for $m$ in equation (II) to find ${\mu }_z$.

    $\begin{array}{c}{\mu }_z=\pm \left[\frac{\left(1.60\times {10}^{-19}\text{ C}\right)\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)}{4\pi \left(9.10\times {10}^{-31}\text{ kg}\right)}\right]\\ =\pm 9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2\end{array}$

Substitute $9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2$ for $\mu$, $10\text{ T}$ for $B$, and $0°$ for $\theta$ in equation (I) to find the energy of electron when the dipole moment due to the spin is aligned parallel to the field.

    $\begin{array}{c}{U}_p=-\left(9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2\right)\left(10\text{ T}\right)\cos 0°\\ =-9.27\times {10}^{-23}\text{ J}\end{array}$

Here, $U_p$ is the energy when dipole moment is parallel to the field.

Substitute $9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2$ for $\mu$, $10\text{ T}$ for $B$, and $180°$ for $\theta$ in equation (I) to find the energy of electron when the dipole moment due to the spin is aligned anti-parallel to the field.

  $\begin{array}{c}{U}_{ap}=-\left(9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2\right)\left(10\text{ T}\right)\cos 180°\\ =-\left(9.27\times {10}^{-24}\text{ A}\cdot {\text{m}}^2\right)\left(10\text{ T}\right)\left(-1\right)\\ =+9.27\times {10}^{-23}\text{ J}\end{array}$

Here, $U_{ap}$ is the energy when dipole moment is anti-parallel to the field.

The energy required to reverse the dipole moment due to the spin of an electron will be the difference between the energies when the dipole moments are parallel and anti-parallel to the field.

Thus, energy required to reverse the direction of the magnetic dipole moment is,

  $U_r=U_{ap}-U_p$        (III)

Here, $U_r$ is the energy required to reverse the direction of the magnetic dipole moment.

Substitute $+9.27\times {10}^{-23}\text{ J}$ for $U_{ap}$, and $-9.27\times {10}^{-23}\text{ J}$ for $U_p$ in equation (III) to find $U_r$.

  $\begin{array}{c}{U}_r=+9.27\times {10}^{-23}\text{ J}-\left(-9.27\times {10}^{-23}\text{ J}\right)\\ =1.85\times {10}^{-22}\text{ J}\\ \text{=}1.85\times {10}^{-22}\text{ J}\times \frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\\ =1.15\times {10}^{-3}\text{ eV}\end{array}$

Therefore, the energy required to reverse the direction of the magnetic dipole moment of an electron due to its spin is $\underset{\_}{1.15\times {10}^{-3}\text{ eV}}$.

(b)

To determine

The ratio of energy required to reverse the direction of the magnetic dipole moment of an electron due to spin to the energy needed to remove an atom from a normal hydrogen atom.

Answer to Problem 18E

The ratio of energy required to reverse the direction of the magnetic dipole moment of an electron due to spin to the energy needed to remove an atom from a normal hydrogen atom is $\underset{\_}{8.45\times {10}^{-5}}$.

Explanation of Solution

The energy needed to remove an atom from a normal hydrogen atom is,

    $U_H=13.6\text{ eV}$

Here, $U_H$ the energy needed to remove an atom from a normal hydrogen atom.

From part (a) the value of $U_r$ is found to be $1.15\times {10}^{-3}\text{ eV}$.

Conclusion:

The ratio of energy required to reverse the direction of the magnetic dipole moment of an electron due to spin to the energy needed to remove an atom from a normal hydrogen atom is,

    $\begin{array}{c}\frac{U_r}{U_H}=\frac{1.15\times {10}^{-3}\text{ eV}}{13.6\text{ eV}}\\ =8.45\times {10}^{-5}\end{array}$

Therefore, the ratio of energy required to reverse the direction of the magnetic dipole moment of an electron due to spin to the energy needed to remove an atom from a normal hydrogen atom is $\underset{\_}{8.45\times {10}^{-5}}$.