# What is the entropy of a hydrogen atom?

What is the entropy of a hydrogen atom, a bound proton and electron?

First attempt: The standard molar entropy of hydrogen gas is 130.68 $$J \, mol^{-1} K^{-1}$$ at $$298 K$$. $$1 \, mol = 6.02214076×10^{23}$$.

Therefore, one hydrogen atom has an entropy of $$2.17 \times 10^{-22} J/K$$.

A thermodynamic definition of entropy is $$S=-k_B \sum_{i} p_i \log{p_i}$$ where $$p_i$$ is the probability of a given microstate and $$k_B$$ is Boltzmann’s constant. This closely resembles the formula for Shannon entropy, $$H=-\sum_i p_i \log{p_i}$$ where $$p_i$$ is the probability of a message $$m_i$$ taken from some message space $$M$$. See info-entropy relationship.

Dividing by $$k_B$$ “yields” the Shannon information, which appears to be 15.72 nats, roughly three bytes. This disagrees with the linked question, which yields a much larger information content.

Second attempt:

The probabilities $$P_i$$ can be obtained from the Boltzmann distribution:

$$P_i = \frac{{e^{-E_i/(k_B T)}}}{{\sum_j e^{-E_j/(k_B T)}}}$$

where $$E_i$$ is the energy of the $$i$$th eigenstate.

The energy levels of a hydrogen atom are given by the formula:

$$E_n = -\frac{{13.6 \, \text{eV}}}{{n^2}}$$

where $$n$$ is the principal quantum number (1, 2, 3, ...).

At $$T = 300 K$$, using the formula above:

1. Calculate the probabilities for each energy eigenstate:

$$P_n = \frac{{e^{-(-13.6 \, \text{eV})/(n^2 k_B \cdot 300 \, \text{K})}}}{{\sum_{j=1}^{\infty} e^{-(-13.6 \, \text{eV})/(j^2 k_B \cdot 300 \, \text{K})}}}$$

2. Calculate the entropy using the formula: $$S = -k_B \sum_{n=1}^{\infty} P_n \log(P_n)$$

The sum in the denominator of $$P_n$$ diverges.

In this paper, the “Shannon entropy” is computed and appears to be something like $$3+\ln(\pi)$$. Any help in understanding where I’m going wrong would be greatly appreciated!

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Aug 5, 2023 at 5:13

"We have to conclude, with some surprise, that the chance of the atom remaining in the ground state or in any other finite state is zero. To make this a little more quantitative, consider an atom in a finite volume, say a sphere of radius $$a$$. Then states extending over a radius much less than $$a$$ will have the same energy as in free space, and states extending much beyond $$a$$ will not exist. Since the mean radius is $$a_0 n^2$$, with $$a_0$$ the Bohr radius, we can find the right order of magnitude by cutting off the sum at $$n = (a/a_0)^{1/2}$$, which, for large $$a$$, gives $$\frac{2}{3}(a/a_0)^{3/2}$$. On the other hand, it is easy to see that the continuous spectrum contributes, in the same limit, an amount proportional to the volume, i.e. to $$a^3$$. So in a really large volume the dominant state for the electron is to be in a state of positive energy; the atom is ionized.