It's a thermodynamics question. Show that at high temperature, such that \( k_B T \gg \hbar \omega \), the partition function of the simple harmonic oscillator is approximately \( Z \approx (\beta \hbar \omega)^{-1} \). Hence find \( U \), \( C \), \( F \) and \( S \) at high temperature. Repeat the problem for the high temperature limit of the rotational energy levels of the diatomic molecule for which \( Z \approx (\beta \hbar^2 / 2I)^{-1} \).

The Hamiltonian for the harmonic oscillator be defined as, H=p22m+12mω2x2=p22m+12kx2The classical partition function be defined as, ZN=1hN∫-∞∞∫-∞∞e-HkBTdxdpFor One SHO:Z1=1h1∫-∞∞∫-∞∞e-HkBTdxdp=1h∫-∞∞dx∫-∞∞e-p22mkBT-x22kBTkdp Here, k be defined as the spring constant while kB is defined as the Boltzmann constant.Re-write the above integral and solve, Z1=1h∫-∞∞dx∫-∞∞e-p22mkBT-x22kBTkdp=1h2πkBT2πkBTk=2πhmkkBT=kBTħω=1βħωZ=βħω-1For N-SHO:ZN=βħω-NTherefore, the internal energy (U) and the specific heat (C) be calculated as, U=-∂∂βlnZ=-∂∂βlnβħω-N=N∂∂βlnβħω=N1βħωħωU=NkBTC=∂U∂TC=∂∂TNkBTC=NkBThe Helmholtz free energy (F) be calculated as, F=-1βlnZ=-1βlnβħω-N=NkBTlnħωkBTThe entropy (S) and the specific heat (C) be calculated as, S=kBlnZ+kBT∂lnZ∂T=kBlnkBTħωN+kBT∂∂TlnkBTħωN=kBlnkBTħωN+NkBTħωkBTkBħωS=NkBlnkBTħω+NkBAccording to the rigid rotator, the energy level of a diatomic molecule be defined as, εrot=JJ+1h28π2I, J=0,1,2,3,....... Where, I be defined as the moment of inertia. J be defined as the rotational quantum number. Therefore, the rotational partition function will be, Zrot=∑Jgrotexp-JJ+1h28π2IkT=∑J2J+1exp-JJ+1h28π2IkT=∫0∞2J+1exp-JJ+1h28π2IkTOn Solving:For high Temperature:Zrot=8π2IkTh2=2Iβħ2=βħ22I-1The quantities U, S, C, F be calculated as, Urot=kT2∂lnZrot∂T=NkTCrot=∂U∂T=NkSrot=klnZrot+kT∂lnZrot∂T=klnβħ22I-N+kT∂∂Tlnβħ22I-N=Nkln2Iβħ2+NkFrot=-1βlnZ=-NkTln2Iβħ2

Answered: Show that at high temperature, such…