Energy of the dust

The probability $p_i$ for an equilibrium system to be in a given energy state is given by

$\displaystyle p_i \sim \exp -E_i/k_BT$ .

For ease of writing, we can say $\gamma_i = E_i/k_BT$ .

The two states states available to a dust particle on the surface are $\displaystyle E = 0, E_1$ .

For the energized state to have a probability $p$ we must have

$\displaystyle p = \frac{e^{-\gamma_1}}{1+e^{-\gamma_1}}$

This is true if we have $\displaystyle \gamma_1 = \ln\frac{1-p}{p}$ .

The energy $E_1$ is then given by $\displaystyle k_BT\ln\frac{1-p}{p}$ .

We know that 20% of the dust particles are in the energized state, so the total energy of all the dust particles is

$\displaystyle E_{total} = \boxed{200E_1=\displaystyle 200\ln\frac{1-0.2}{0.2}\approx 277 k_BT}$

Using the canonical ensemble representation of the system, we know that the probability of each energy state is given by:

$p_i = \frac{1}{Z} e^{-\frac{E_i}{kT_0}}$

Where $Z$ is just the normalizing constant:

$Z = \sum_i p_i$

Using this information and the fact that $E_0 = 0$ , we can see that the probability of being in energy state $E_1$ is:

$p_1 = \frac{\frac{1}{Z} e^{-\frac{E_1}{kT_0}}}{1 + \frac{1}{Z} e^{-\frac{E_1}{kT_0}}}$

With a bit of algebra after replacing $p_1 with 0.2$ we get:

$\frac{E_1}{kT_0} = -ln(0.25) \approx 1.38629$

Since 20% of the 1000 particles are in energy state 1 and the other particles have energy 0, we just need to multiply the above value by 200 to obtain the final answer.

$p_1=0.2 \mbox{ for } E_1$

$p_0=0.8 \mbox{ for } E_0$

Boltzmann

$p_1/p_0=e^{(E_0-E_1)/k_BT}$

$-\ln{(0.2/0.8)}=-\ln{0.25}=1.386.. = E_1/k_BT$

$sum= 200*1.386..=277.26..$