To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.
Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by
[tex]z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}[/tex]
Where,
[tex]\epsilon_i =[/tex] energy of that state
k = Boltzmann's constant
T = Temperature
With our values we have that
T= 250K
[tex]k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}[/tex]
[tex]\epsilon_1=0J[/tex]
[tex]\epsilon_2=1.6*10^{-21}J[/tex]
[tex]\epsilon_3=1.6*10^{-21}J[/tex]
To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as
[tex]\beta = \frac{1}{kT}[/tex]
[tex]\beta = \frac{1}{(1.381*10^{23} m^2)(250)}[/tex]
[tex]\beta = 2.9*10^{20}J[/tex]
Therefore the average energy would be,
[tex]\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}[/tex]
Replacing with our values we have
[tex]\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}[/tex]
[tex]\bar{\epsilon} = 0.9*10^{-22}J[/tex]
Therefore the average internal energy is [tex]\bar{\epsilon} = 0.9*10^{-22}J[/tex]
