Question

A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment with temperature 250 K. 1) What is the average internal energy?

Answer 1

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

`z = \sum\limit_i e^{-(\epsilon_i)/(K_0T)}`

Where,

`\epsilon_i =` energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

`k = 1.381*10^(23) m^2 kg s^(-2) K^(-1)`

`\epsilon_1=0J`

`\epsilon_2=1.6*10^(-21)J`

`\epsilon_3=1.6*10^(-21)J`

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

`\beta = (1)/(kT)`

`\beta = (1)/((1.381*10^(23) m^2)(250))`

`\beta = 2.9*10^(20)J`

Therefore the average energy would be,

`\bar{\epsilon} =(\sum \epsilon_i e^(-\beta \epsilon_i))/(\sum e^(-\beta \epsilon_i))`

Replacing with our values we have

`\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)}}`

`\bar{\epsilon} = 0.9*10^(-22)J`

Therefore the average internal energy is
`\bar{\epsilon} = 0.9*10^(-22)J`