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 4 × 10 − 21 J apart. (a) When the nanoparticle’s energy is in the range 5 × 4 × 10 − 21 J to 6 × 4 × 10 − 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 Kelvin. (b) When the nanoparticle’s energy is in the range 8 × 4 × 10 − 21 J to 9 × 4 × 10 − 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 × 4 × 10 − 21 J to 9 × 4 × 10 − 21 J, what is the approximate heat capacity per atom?

Answer 1

Approximate temperature T when energy is in between 5 and 6 quanta is 1 T = slope of the entropy versus energy graph. At q = 5: To find the slope, we have 1 T = Change in entropy Change in energy = (0.159 − 0.141) × 10−21 J/K (6 − 5) × 4 × 10−21 J = 4.5 × 10−3 1/K ⇒ T = 222.2 K.