Boltzmann Sample Exercises Using the TI-85, TI-86, or TI-89 Equation Solver

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Full solutions are not given here. Instead, I just show how the equation gets put into the TI-85 and solved.

A quantum system has 6.022 x 1023 particles in the ground state. If the temperature is 300 K and a the energy levels are evenly spaced with a separation of 1 x 10−20 J, how many particles would be expected to be in the second level? The third level?

eqn:nj/ni=e^(-dEij/(kb*T))
 nj=                Press [SOLVE]==>     .nj=1.610505521378E22
 ni=6.022E23                              ni=6.022E23
 dEij=1E-20                               dEij=1E-20
 kb=1.38066E-23                           kb=1.38066E-23
 T=200                                    T=200
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: There would be 1.611 x 1022 particles in the second energy level.

Now we put this value in for ni and solve again.

eqn:nj/ni=e^(-dEij/(kb*T))
 nj=                Press [SOLVE]==>     .nj=4.30708740350E20
 ni=1.611E22                              ni=1.611E22
 dEij=1E-20                               dEij=2E-20
 kb=1.38066E-23                           kb=1.38066E-23
 T=200                                    T=200
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: There would be 4.31 x 1020 particles in the third energy level.

At what temperature would a quantum system having evenly spaced energy levels with a separation of 1 x 10−20 J show a ratio of 1/1000 for number of particles in adjacent levels? What would the temperature have to be if there were just 1 in a million particles excited?

Even though we only know the ratio of nj/ni, we can assume there are 1000 particles in the lower level and just put 1 in for the number of particles in the upper level.

eqn:nj/ni=e^(-dEij/(kb*T))
 nj=1                                     nj=1
 ni=1000                                  ni=1000
 dEij=1E-20                               dEij=1E-20
 kb=1.38066E-23                           kb=1.38066E-23
 T=                 Press [SOLVE]==>     .T=104.85190220697632
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: The temperature would have to be 105 Kelvin for nj/ni to be 1/1000.

Now we change ni to 100000 and solve again.

eqn:nj/ni=e^(-dEij/(kb*T))
 nj=1                                     nj=1
 ni=1000000                               ni=1000000
 dEij=1E-20                               dEij=1E-20
 kb=1.38066E-23                           kb=1.38066E-23
 T=104.85190220697632  Press [F5]==>     .T=52.42595110348816
 bound={-1E99,1E99}                       bound={-1E99,1E99}
Answer: The temperature would now be 52 Kelvin.