Question

Consider 6 identical, non-interacting, spin $\dfrac{1}{2}$ atoms arranged on a crystal lattice at absolute temperature $T$. The z-component of the magnetic moment of each of these atoms can be $\pm \mu_B$. If $P$ and $Q$ are the probabilities of the net magnetic moment of the solid being $2\mu_B$ and $6\mu_B$ respectively, what is the value of $\dfrac{P}{Q}$ (in integer)?

• A. 15
• B. 10
• C. 20
• D. 25

Hint:

When dealing with spin-$\dfrac{1}{2}$ particles in a crystal lattice, the number of possible configurations can be found by counting the different ways the particles can align to give the desired net magnetic moment.

Answer 1

The Correct Option is A

For a system of 6 non-interacting spin $\dfrac{1}{2}$ particles, the net magnetic moment is given by the sum of individual magnetic moments, each of which can be either $\pm \mu_B$. The probabilities $P$ and $Q$ are associated with specific values of the net magnetic moment. For the net magnetic moment to be $2\mu_B$, we need 2 particles with spin-up and 4 particles with spin-down. Similarly, for the net magnetic moment to be $6\mu_B$, all 6 particles must be in the spin-up state. The ratio of $P$ and $Q$ is calculated as: \[ \frac{P}{Q} = \frac{\text{probability of 2 spin-up and 4 spin-down}}{\text{probability of 6 spin-up}} = 15. \] Thus, the correct answer is (A) 15.