Problem Statement
Two energy states are separated by $\Delta E = 0.020\ \text{eV}$. Find the ratio of populations $N_2/N_1$ at $T=300\ \text{K}$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Boltzmann distribution: $N_i \propto e^{-E_i/k_BT}$, so:
Step 2 — Apply the relevant physical law or equation: $$\frac{N_2}{N_1} = e^{-\Delta E/k_BT}$$
$$\Delta E = 0.020\times1.6\times10^{-19} = 3.2\times10^{-21}\ \text{J}$$
$$\frac{\Delta E}{k_BT} = \frac{3.2\times10^{-21}}{1.38\times10^{-23}\times300} = \frac{3.2\times10^{-21}}{4.14\times10^{-21}} = 0.773$$
$$\frac{N_2}{N_1} = e^{-0.773} \approx 0.461$$
Step 3 — Solve algebraically for the unknown: Result: $N_2/N_1 \approx 0.46$ — slightly fewer in the higher state, as expected for $k_BT > \Delta E$.
Worked Calculation
$$\frac{N_2}{N_1} = e^{-\Delta E/k_BT}$$
$$\Delta E = 0.020\times1.6\times10^{-19} = 3.2\times10^{-21}\ \text{J}$$
$$\frac{\Delta E}{k_BT} = \frac{3.2\times10^{-21}}{1.38\times10^{-23}\times300} = \frac{3.2\times10^{-21}}{4.14\times10^{-21}} = 0.773$$
Boltzmann distribution: $N_i \propto e^{-E_i/k_BT}$, so:
$$\frac{N_2}{N_1} = e^{-\Delta E/k_BT}$$
$$\Delta E = 0.020\times1.6\times10^{-19} = 3.2\times10^{-21}\ \text{J}$$
$$\frac{\Delta E}{k_BT} = \frac{3.2\times10^{-21}}{1.38\times10^{-23}\times300} = \frac{3.2\times10^{-21}}{4.14\times10^{-21}} = 0.773$$
$$\frac{N_2}{N_1} = e^{-0.773} \approx 0.461$$
Result: $N_2/N_1 \approx 0.46$ — slightly fewer in the higher state, as expected for $k_BT > \Delta E$.
Answer
$$\boxed{\frac{N_2}{N_1} = e^{-0.773} \approx 0.461}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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