## 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.