Question

When a magnetic dipole is placed in a magnetic field, it has a natural tendency to minimize its potential energy by aligning itself with the field. If there is sufficient thermal energy present, however, the dipole may rotate so that it is no longer aligned with the field. Using \(k_{\mathrm{B}} T\) as a measure of the thermal energy, where \(k_{B}\) is Boltzmann's constant and \(T\) is the temperature in kelvins, determine the temperature at which there is sufficient thermal energy to rotate the magnetic dipole associated with a hydrogen atom from an orientation parallel to an applied magnetic field to one that is antiparallel to the applied field. Assume that the strength of the field is \(0.15 \mathrm{~T}\).

Short answer

The temperature is approximately 0.66 K.

Step-by-step solution

Calculate potential energy difference

The potential energy of a magnetic dipole in a magnetic field is given by \(U = -\boldsymbol{\mu} \cdot \textbf{B}\), where \(\boldsymbol{\mu}\) is the magnetic moment of the dipole and \(\textbf{B}\) is the magnetic field. Therefore, the potential energy difference between antiparallel (\(180^\circ\), or \(\pi\) radians) and parallel (\(0^\circ\), or \(0\) radians) orientations is given by \(\Delta U = U_{\pi} - U_0\). We know that for a hydrogen atom, the magnetic moment \(\boldsymbol{\mu} = -\frac{e \hbar}{2m_e} \textbf{M}\), where e is the elementary charge, \(m_e\) is the electron mass, \(\hbar\) is the reduced Planck constant, and \(\textbf{M}\) is the magnetic quantum number of the electron. Since \(\Delta U = -\boldsymbol{\mu} \cdot \Delta \textbf{B}\), we can calculate \(\Delta U\) by finding the difference between the magnetic field in antiparallel and parallel orientations. We are given that the magnetic field strength is \(0.15 \mathrm{~T}\), so the potential energy difference can be expressed as follows: \(\Delta U = -\frac{e \hbar}{2m_e} (\textbf{B}(\pi) - \textbf{B}(0)) = -\frac{e \hbar}{2m_e} (2\textbf{B})\)

Equate potential energy difference to thermal energy

Now that we have the potential energy difference, we must equate it to the thermal energy, given by \(k_{\mathrm{B}} T\), to find the temperature at which there is sufficient thermal energy to rotate the magnetic dipole: \(\Delta U = k_{\mathrm{B}} T\) Substituting the expression for \(\Delta U\): \(-\frac{e \hbar}{m_e} \textbf{B} = k_{\mathrm{B}} T\)

Solve for the temperature T

Now, let's solve the equation for T: \(T = \frac{e \hbar}{m_e k_{\mathrm{B}}} \textbf{B}\) Using the given values and known constants: \(T = \frac{(1.6 \times 10^{-19} \mathrm{C})(1.054 \times 10^{-34} \mathrm{J\cdot s})}{(9.11 \times 10^{-31} \mathrm{kg})(1.38 \times 10^{-23} \mathrm{J/K})} \times 0.15 \mathrm{T}\) Finally, we can compute the temperature: \(T \approx 0.66 \mathrm{K}\) Therefore, the temperature at which there is sufficient thermal energy to rotate the magnetic dipole associated with a hydrogen atom from an orientation parallel to an applied magnetic field with a strength \(0.15 \mathrm{~T}\) to one that is antiparallel to the applied field is approximately 0.66 K.