Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 2 / A6 / b

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

“Low” and “high” compare $kT$ with the vibrational spacing $hc\tilde{\omega }$, not temperature with an arbitrary number of kelvin. For $T\ll {\theta }_v$, occupation of excited levels is exponentially suppressed andThis vanishing result uses the ground-relative energy zero. For $T\gg {\theta }_v$, expand $e^{{\theta }_v/T}-1\sim {\theta }_v/T$ to obtainThis agrees with equipartition: the oscillator has one quadratic kinetic and one quadratic potential term, contributing $kT/2$ each. The corresponding limiting vibrational heat capacity is zero at low temperature and $Nk$ at high temperature.