THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

446 16. UltrasonicsPlanck constant 1.055 ×10 −34 J s. Let us assume that J is the most probable valueaccording to the Boltzmann distribution. The value of 2(J+ 1)B for a typicalmolecule (for example, O 2 ) in temperature units is about 1 K. This indicates thatin a gas above 1 K all collisions will have enough translational energy to engendermultiple changes, with the result that rotation rapidly equilibrates with translation.Hydrogen, however, constitutes an exception, because it has much larger rotationalenergy level spacing due to its small moment of inertia. According to Winter andHill (1967), as many as 350 collisions may be necessary to transfer a quantum of rotationalenergy in the hydrogen molecule. As a pressure of 1 atm, this correspondsto a relaxation time of 2 × 10 −8 s. It should be understood that a specific collisioneither does or does not transfer a quantum of rotational energy in the hydrogenmolecule. The 350-collision average indicates that only one of the 350 collisionspossesses the proper geometry and energy to execute a transfer of one quantum ofrotational energy. The number of collisions necessary, on the average, to engenderthe transfer one quantum of rotational energy is termed the collision number Z.Where rotational energy is entailed, the collision number is written as Z rot . Theinverse of this dimensionless parameter represents the probability of transferringa quantum in a collision, symbolized by P rot . Rotational energy levels are spacedunevenly, i.e., a 1 → 2 transition should be more probable than a 2 → 3 transition.orProt 2→3 .As a rule, the probability for transferring a quantum of energy through a collisiondrops off rapidly with the size of the quantum transferred. Because vibrationalenergy levels are much more widely spaced than rotational energy levels,the vibrational relaxation times are considerably longer than rotational relaxationtimes. Vibrational levels in single vibrational mode are virtually evenly spaced,so energy can therefore be exchanged between levels (e.g., the vibrational quantumnumber goes up in one molecule and goes down in another) with hardlyany energy exchanged between vibration and translational. Thus, the vibration-tovibrationexchanges occur very rapidly. The amount of time it takes for energy totransfer between translation and the lowest lying vibrational level determines thevibrational relaxation time.Because this energy level varies greatly for different molecules, the probabilityof vibrational energy transfer during a collision also varies greatly. In the case ofN 1 molecules colliding with N 2 molecules, Z 10 (the number of collisions needed totransfer energy from the lowest vibrational energy to translation) is approximately1.5 × 10 11 , so the relaxation time is close to 15 s (Zuckerwar and Griffin, 1980).Larger molecules possess very closely spaced vibrational energy levels, so fewercollisions are required to transfer a quantum of first-level vibrational energy intotranslational energy. For example, a relatively large molecule such as C 2 H 6 needsto undergo about 100 collisions to execute this type of transfer.Let us consider the case of a vibrational state that is excited to an energy E v (thesubscript v denotes vibration) that is greater than energy E v (T tr ) that would existin a Boltzmann equilibrium with translation. This excess vibrational energy willequilibrate with translational energy, in accordance with the following standardThe probability of these events occurring is distinguished by the symbolsP 1→2rot