gen-define-the-boltzmann-distribution-and-give-one-short-worked-0ee1

Define the Boltzmann distribution and give one short worked example.

# Boltzmann Distribution ## Definition The **Boltzmann distribution** gives the probability that a system in thermal equilibrium at temperature T will be found in a state with energy E: $$P(E) = \frac{e^{-E/k_BT}}{Z}$$ where: - $k_B$ = Boltzmann constant (1.381 × 10⁻²³ J/K) - $T$ = absolute temperature (K) - $Z$ = partition function = $\sum_i e^{-E_i/k_BT}$ (normalization constant) **Key insight**: States with lower energy are exponentially more probable than states with higher energy. ## Worked Example **Problem**: A two-level system has energy levels E₀ = 0 and E₁ = 0.025 eV. At T = 300 K, what fraction of particles are in each state? **Solution**: First, convert temperature to energy units: $k_BT = (8.617 \times 10^{-5} \text{ eV/K})(300 \text{ K}) = 0.0259$ eV Calculate partition function: $$Z = e^{-E_0/k_BT} + e^{-E_1/k_BT} = e^{0} + e^{-0.025/0.0259} = 1 + e^{-0.965} = 1 + 0.381 = 1.381$$ Calculate probabilities: $$P(E_0) = \frac{e^{0}}{1.381} = \frac{1}{1.381} = 0.724 = 72.4\%$$ $$P(E_1) = \frac{e^{-0.965}}{1.381} = \frac{0.381}{1.381} = 0.276 = 27.6\%$$ **Result**: About 72% of particles occupy the ground state, 28% the excited state.