Let's solve the problem step by step: ### **Problem Restatement** A system of \( N \) distinguishable particles, each with a magnetic moment \( \mu \), can occupy two levels with energies: - Upper: \( +\mu x_0/2 \) - Lower: \( -\mu x_0/2 \) when the magnetic field intensity is \( x_0 \). - **Upper level:** Magnetic moment antiparallel to field - **Lower level:** Magnetic moment aligned parallel to field The system is prepared so that **1/3 of the particles are in the upper level** and is isolated. --- ### **(a) Find the energy and net magnetic moment of the system.** Let: - \( N_1 \) = Number in upper level = \( N/3 \) - \( N_2 \) = Number in lower level = \( 2N/3 \) **Total energy:** \[ E = N_1 \left(\frac{\mu x_0}{2}\right) + N_2 \left(-\frac{\mu x_0}{2}\right) \] \[ E = \frac{N}{3}\left(\frac{\mu x_0}{2}\right) + \frac{2N}{3}\left(-\frac{\mu x_0}{2}\right) \] \[ E = \frac{N\mu x_0}{6} - \frac{2N\mu x_0}{6} = -\frac{N\mu x_0}{6} \] **Net magnetic moment:** Each particle: - Upper level: \( -\mu \) (antiparallel) - Lower level: \( +\mu \) (parallel) \[ \mathcal{M} = N_1 (-\mu) + N_2 (+\mu) \] \[ \mathcal{M} = \frac{N}{3}(-\mu) + \frac{2N}{3}(+\mu) \] \[ \mathcal{M} = -\frac{N\mu}{3} + \frac{2N\mu}{3} = \frac{N\mu}{3} \] --- ### **(b) Calculate the change of the energy and change of net magnetic moment if the magnetic intensity is reduced to \( x_0/2 \) (system isolated)** - **Isolated:** Populations \( N_1 \) and \( N_2 \) do **not change**. Let \( x = x_0/2 \): Energy: \[ E' = N_1 \left(\frac{\mu x}{2}\right) + N_2 \left(-\frac{\mu x}{2}\right) \] \[ E' = \frac{N}{3}\left(\frac{\mu x_0}{4}\right) + \frac{2N}{3}\left(-\frac{\mu x_0}{4}\right) \] \[ E' = \frac{N\mu x_0}{12} - \frac{2N\mu x_0}{12} = -\frac{N\mu x_0}{12} \] Change in energy: \[ \Delta E = E' - E = -\frac{N\mu x_0}{12} - \left(-\frac{N\mu x_0}{6}\right) \] \[ \Delta E = -\frac{N\mu x_0}{12} + \frac{2N\mu x_0}{12} = \frac{N\mu x_0}{12} \] Net magnetic moment: \[ \mathcal{M}' = \mathcal{M} = \frac{N\mu}{3} \] \[ \Delta \mathcal{M} = 0 \] --- ### **(c) Change in net magnetic moment if magnetic intensity is reduced to \( x_0/2 \) but energy stays constant** Now, energy remains at \( E = -\frac{N\mu x_0}{6} \), but the energy levels are: - Upper: \( +\mu x_0/4 \) - Lower: \( -\mu x_0/4 \) Let \( N_1' \) be the new number in the upper level. \[ E = N_1' \left(\frac{\mu x_0}{4}\right) + (N - N_1')\left(-\frac{\mu x_0}{4}\right) \] \[ E = \frac{\mu x_0}{4}N_1' - \frac{\mu x_0}{4}(N - N_1') = \frac{\mu x_0}{4}(2N_1' - N) \] Set \( E = -\frac{N\mu x_0}{6} \): \[ \frac{\mu x_0}{4}(2N_1' - N) = -\frac{N\mu x_0}{6} \] \[ 2N_1' - N = -\frac{2N}{6} = -\frac{N}{3} \] \[ 2N_1' = N - \frac{N}{3} = \frac{2N}{3} \] \[ N_1' = \frac{N}{3} \] **So, the populations don't change.** \[ \mathcal{M}' = N_1'(-\mu) + (N-N_1')(+\mu) = \frac{N}{3}(-\mu) + \frac{2N}{3}(+\mu) = \frac{N\mu}{3} \] So, \[ \Delta \mathcal{M} = 0 \] --- ## **Final Answers** ### (a) - **Total energy:** \(\boxed{-\frac{N\mu x_0}{6}}\) - **Net magnetic moment:** \(\boxed{\frac{N\mu}{3}}\) ### (b) - **Change in energy:** \(\boxed{+\frac{N\mu x_0}{12}}\) - **Change in net magnetic moment:** \(\boxed{0}\) ### (c) - **Change in net magnetic moment:** \(\boxed{0}\)