Chapter 29, Problem 24E

(a)

To determine

The energy of the proton in the magnetic field of $1.2\text{T}$ if the z component of the spin angular momentum is along the field.

Answer to Problem 24E

The energy of the proton when the z component of the spin angular momentum is along the magnetic field is $-16.90\times {10}^{-27}\text{J}$.

Explanation of Solution

When the charge spins, it has the magnetic dipole moment $\mu$, associated with it which is proportional to the spin angular momentum represented by $S$.

Write the expression for the magnetic dipole moment.

  $\mu =2.79\frac{eS}{m_p}$        (I)

Here, $\mu$ is the magnetic dipole moment, $e$ is the charge, $S$ is the spin angular momentum and $m_p$ is the mass of the proton.

Write the expression for the energy of the dipole.

  $U=-\mu B_T$        (II)

Here, $U$ is the energy of the dipole and $B_T$ is the magnetic field.

The value of the spin angular momentum is $+\frac{\hslash }{2}$ or $\frac{-\hslash }{2}$ for the proton.

Conclusion:

Substitute $+\frac{\hslash }{2}$ for $S$ in equation (I).

  $\mu =2.79\frac{e\hslash }{2m_p}$                                                                                     

Substitute ${\mu }_n$ for $\frac{e\hslash }{2m_p}$ in above equation

  $\mu =2.79{\mu }_n$        (III)

Here, ${\mu }_n$ is the nuclear magnetron.

Substitute $5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}$ for ${\mu }_n$ in equation (III).

  $\mu =2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)$

Substitute $2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)$ for $\mu$ and $1.2\text{T}$ for $B_T$ in equation (II).

$\begin{array}{l}U=-2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)1.2\text{T}\\ U=-16.90\times {10}^{-27}\text{J}\end{array}$

Thus, the energy of the proton when the z component of the spin angular momentum is along the magnetic field is $-16.90\times {10}^{-27}\text{J}$.

(b)

To determine

The energy of the proton in the magnetic field of $1.2\text{T}$ if the z component of the spin angular momentum is opposite to the field.

Answer to Problem 24E

The energy of the proton when the z component of the spin angular momentum is along the magnetic field is $16.90\times {10}^{-27}\text{J}$.

Explanation of Solution

When the charge spins, it has the magnetic dipole moment $\mu$, associated with it which is proportional to the spin angular momentum represented by $S$.

Write the expression for the magnetic dipole moment.

  $\mu =2.79\frac{eS}{m_p}$        (I)

Here, $\mu$ is the magnetic dipole moment, $e$ is the charge, $S$ is the spin angular momentum and $m_p$ is the mass of the proton.

Write the expression for the energy of the dipole.

  $U=-\mu B_T$        (II)

Here, $U$ is the energy of the dipole and $B_T$ is the magnetic field.

The value of the spin angular momentum is $+\frac{\hslash }{2}$ or $\frac{-\hslash }{2}$ for the proton.

Substitute $-\frac{\hslash }{2}$ for $S$ in equation (I).

  $\mu =-2.79\frac{e\hslash }{2m_p}$        (III)

Write the expression for the nuclear magnetron.

  ${\mu }_n=\frac{e\hslash }{2m_p}$                                                                                            

Here, ${\mu }_n$ is the nuclear magnetron.

Conclusion:

Substitute ${\mu }_n$ for $\frac{e\hslash }{2m_p}$ in equation (3).

  $\mu =-2.79{\mu }_n$        (IV)

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$ , $1.05\times {10}^{-34}\text{J}\text{.s}$ for $\hslash$ and $1.6726\times {10}^{-27}\text{kg}$ for $m_p$ in the above equation.

  $\begin{array}{l}{\mu }_n=\frac{1.6\times {10}^{-19}\text{C}\left(1.05\times {10}^{-34}\text{J}\text{.s}\right)}{2\left(1.6726\times {10}^{-27}\text{kg}\right)}\\ {\mu }_n=5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\end{array}$

Substitute $5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}$ for ${\mu }_n$ in equation (IV).

  $\mu =-2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)$

Substitute $2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)$ for $\mu$ and $1.2\text{T}$ for $B_T$ in equation (II).

$\begin{array}{l}U=2.79\left(5.05\times {10}^{-27}\text{A}{\text{.m}}^{\text{2}}\right)1.2\text{T}\\ U=16.90\times {10}^{-27}\text{J}\end{array}$

Thus, the energy of the proton when the z component of the spin angular momentum is along the magnetic field is $16.90\times {10}^{-27}\text{J}$.