# What units do I use for energy in the Boltzmann distribution?

I am a mathematics student doing a course in thermodynamics and am working on the basics now. The question I am asked is:

Suppose two states differ in energy by one electronvolt. What can be said about the ratio of the population at $$T = \pu{300 K}$$?

Now it is clear that I will use the formula:

$$\frac{N_i}{N_j} = \exp\left(-\frac{E_i-E_j}{kT}\right)$$

where $$k = \pu{1.381 \times 10^−23 J K-1}$$. Now if let $$E_i$$ be the higher energy state, then am I right in saying that $$E_i-E_j = \pu{1 eV}$$, or should I convert this to some other unit?

## 1 Answer

Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that $$kT$$ has to have the same units as $$E_i-E_j$$, so the terms in the exponent cancel. If you're measuring temperature in $$\pu{K}$$ and Boltzmann's constant in $$\pu{J/K}$$, then the units of energy you have to use are joules, $$\pu{J}$$.

• Alternatuvely, you can convert $k_B$ to using $\mathrm{eV \cdot K^{-1}}$. Commented May 9, 2019 at 17:39