# What is $k_B$ in the context of this question?

1000 atoms are in equilibrium at temperature T. Each atom has two energy states, $E_1$ and $E_2$, where $E_2 > E_1$ . On average, there are 200 atoms in the higher energy state and 800 in the lower energy state.

What is the Temperature?

Why is the answer $T = (E_2 - E_1)/(2 k_B ln 2)$, specifically what is $k_B$?

The original question can be found here under Question 3.

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You already accepted an answer, so I guess the solution is clear now. However, I don't see how you get rid of $P$ and $Z$ in the formula Emilio Pisanty posted without using the fact that the partition is described by a biominal distribution and so $\frac{p_1}{p_2}=\frac{N_1^\text{max}}{N^\text{tot}-N_1^\text{max}}=4$. –  NikolajK Jul 22 '12 at 20:57

The symbol $k_\textrm{B}$ pretty much invariably denotes Boltzmann's constant.
Apart from that, your question is asking about the statistical mechanics fact that for a canonical ensemble (i.e. a physical system in contact with a heat bath at some temperature $T$) the probability for the system to have energy $E$ is equal to $$P = \frac{1}{Z}e^{-E/k_\textrm{B}T}$$ where $Z$ is a normalization factor known as a partition function.