A nanoparticle containing 6 atoms can be modeled approximately as an Einstein so
ID: 1566078 • Letter: A
Question
A nanoparticle containing 6 atoms can be modeled approximately as an Einstein solid of 18 independent oscillators. The evenly spaced energy levels of each oscillator are 4e-21 J apart. Use k = 1.4e-23 J/K. a) When the nanoparticle's energy is in the range 5(4e-21) J to 60(4e-21) J, what is the approximate temperature? (In order to keep precision for calculating the heat capacity, give the result to the nearest tenth of a degree.) b) When the nanoparticle's energy is in the range 8(4e-21) J to 9(4e-21) J, what is the approximate temperature? (In order to keep precision for calculating the heat capacity, give the result to the nearest tenth of a degree.) (c) When the nanoparticle's energy is in the range 5(4e-21) J to 9(4e-21) J, what is the approximate heat capacity per atom? Note that between parts (a) and (b) the average energy increased from "5.5 quanta" to "8.5 quanta". As a check, compare your result with the high temperature limit of 3k, where k = 1.4e-23 J/K.Explanation / Answer
a) 1/T = change in entropy / change in energy
= kb*[ln(100947) - ln(26334) ] / (6 - 5) * 4 *10^-21
= (1.4e-23)*[ln(100947) - ln(26334) ] / (6 - 5) * 4 *10^-21
= 4.7 *10^-3 J/K
==> T = 212.63 K
b) 1/T = change in entropy / change in energy
= kb*[ln(3124550) - ln(1081575) ] / (9 - 8) * 4 *10^-21
= (1.4e-23)*[ln(3124550) - ln(1081575) ] / (9 - 8) * 4 *10^-21
= 3.71*10^-3 J/K
==> T = 269.32 K
c) To calculate the heat capacity per atom, one needs to find the energy difference between the points
a and b and the corresponding change in temperature.
Ta = 269.32 K and Tb = 212.63K
Delta T = Tb - Ta = 269.32 - 212.63 = 56.69 K
Change in energy = 4 *10^-21 J
heat capacity per atom C = 4 *10^-21 / 2*56.69
= 3.5*10^-23 J/K
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