Find the minimum torque magnitude \(| \vec{\tau}\) | that acts on the orbital magnetic dipole of a \(3 p\) electron in an external magnetic field of \(2.50 \times 10^{-3} \mathrm{T}\)

First, we need to determine the quantum numbers of the 3p electron. For the 3p orbital, the following quantum numbers apply: Principal quantum number (n): \(n = 3\) Orbital angular momentum quantum number (l): \(l = 1\), since it is a p orbital (0 corresponds to s, 1 to p, 2 to d, and so on) Magnetic quantum number (m): The magnetic quantum number (m) can range from \(-l\) to \(+l\), so in this case, it can be -1, 0, or 1. To find the minimum torque, we need to choose the alignment which results in the smallest angle between the magnetic dipole moment and the external magnetic field. This occurs when \(m = 1\), since it will be aligned most closely with the external magnetic field.

The magnetic dipole moment (\(\vec{\mu}\)) is related to the angular momentum (\(\vec{L}\)) of the electron by the equation: \(\vec{\mu} = -g \mu_{B} \frac{\vec{L}}{\hbar}\) Here, \(g\) is the Landé g-factor (approximately 1 for orbital angular momenta), \(\mu_B\) is the Bohr magneton and \(\hbar\) is the reduced Planck constant. The magnitude of the angular momentum is given by: \(|\vec{L}| = \sqrt{l(l + 1)} \hbar\) Substitute the value of \(l\) into this equation to find the magnitude of the angular momentum of the 3p electron: \(|\vec{L}| = \sqrt{1(1 + 1)} \hbar = \sqrt{2} \hbar\) Now, we can determine the magnetic dipole moment's magnitude: \(|\vec{\mu}| = |-g \mu_{B} \frac{\vec{L}}{\hbar} | = |-1 \cdot\mu_{B} \sqrt{2} | = \sqrt{2} \mu_B\)

Now that we know the magnetic dipole moment magnitude, we can find the minimum torque magnitude using the following equation: \(|\vec{\tau}| = |\vec{\mu}| \times |\vec{B}| \cdot \sin\theta\) The torque will be minimum when the angle \(\theta\) between \(\vec{\mu}\) and \(\vec{B}\) is minimum. That is, when the magnetic quantum number \(m = 1\), the smallest possible angle is \(\theta = 0^\circ\). Due to spatial quantization of the electron, the magnetic dipole moment can only take discrete values for the angle, so the minimum nonzero angle, in this case, is \(90^\circ\). Therefore: \(|\vec{\tau}| = \sqrt{2} \mu_B \times 2.50 \times 10^{-3} \mathrm{T} \times \sin(90^\circ)\) The sine of \(90^\circ\) is 1, so: \(|\vec{\tau}| = \sqrt{2} \mu_B \times 2.50 \times 10^{-3} \mathrm{T}\)

To compute the minimum torque magnitude, we substitute the value of the Bohr magneton \(\mu_B = 9.274 \times 10^{-24}\) J/T: \(|\vec{\tau}| = \sqrt{2} \times (9.274 \times 10^{-24} \text{J/T}) \times 2.50 \times 10^{-3} \mathrm{T}\) \(|\vec{\tau}| = 3.29 \times 10^{-26} \mathrm{Nm}\) So the minimum torque magnitude acting on the orbital magnetic dipole of a 3p electron in an external magnetic field of \(2.50 \times 10^{-3} \mathrm{T}\) is \(3.29 \times 10^{-26} \mathrm{Nm}\).