Question

Consider \(N\) classical particles at temperature \(T\), each of which can have two possible energies 0 and \(\epsilon\). The number of particles in the lower energy level (\(N_0\)) and higher energy level (\(N_\epsilon\)) are related by (\(k_B\) is the Boltzmann constant):

• A. \(\dfrac{N_0}{N_\epsilon} = e^{-\frac{\epsilon}{k_B T}}\)
• B. \(\dfrac{N_0}{N_\epsilon} = e^{\frac{\epsilon}{k_B T}}\)
• C. \(\dfrac{N_0}{N_\epsilon} = 1 + e^{\frac{\epsilon}{k_B T}}\)
• D. \(\dfrac{N_0}{N_\epsilon} = 1 - e^{-\frac{\epsilon}{k_B T}}\)

Hint:

In thermal equilibrium, population ratio between two levels depends exponentially on the energy difference and temperature.

Answer 1

The Correct Option is B

Step 1: Boltzmann distribution ratio.
The relative population of two energy levels is given by \[ \frac{N_\epsilon}{N_0} = e^{-\frac{(\epsilon - 0)}{k_B T}} = e^{-\frac{\epsilon}{k_B T}} \]

Step 2: Take reciprocal.
\[ \frac{N_0}{N_\epsilon} = e^{\frac{\epsilon}{k_B T}} \]

Step 3: Conclusion.
Hence, the population ratio between the ground and excited state is \(\dfrac{N_0}{N_\epsilon} = e^{\frac{\epsilon}{k_B T}}\).