The energy levels of a three-dimensional rigid rotor of moment of inertial I are given by EJ, M=

step 1 In this question, we have a rigid rotor. So this rigid rotor is made up of a messless rod with t

The energy levels of a three-dimensional rigid rotor of...

Key Concepts

Quantum Rigid Rotor

The quantum rigid rotor is a model used to describe the rotation of a molecule or other rigid body in three dimensions. Its energy levels are quantized and characterized by the quantum number J, with each level possessing a degeneracy (multiplicity) related to the magnetic quantum number M. This model provides the basis for understanding rotational spectra and the quantization of angular momentum in molecular systems.

Boltzmann Statistics

Boltzmann statistics is a framework within statistical mechanics that assigns probabilities to the occupancy of energy states in a system at thermal equilibrium. It involves the use of the Boltzmann factor, exp(–?E), where ? is the inverse temperature (1/kT) and E is the energy of a state. This approach is fundamental for calculating thermodynamic quantities, such as the internal energy, by summing over all possible states weighted by their energetic factors.

Partition Function

The partition function is a central quantity in statistical mechanics that encapsulates all the statistical properties of a system. For a system described by a given set of energy levels, the partition function is the sum over states of the Boltzmann factors, often including degeneracy factors. Once the partition function is known, macroscopic observables such as internal energy and specific heat can be derived through differentiation with respect to temperature.

High Temperature/Integral Approximation

At high temperatures, the spacing between quantized energy levels becomes small compared to the thermal energy kT, allowing the discrete sum over quantum states to be approximated by a continuous integral. This semi-classical approximation simplifies the calculation of thermodynamic properties such as the internal energy and specific heat by replacing the sum over quantum numbers with an integral, which is easier to evaluate analytically.

Specific Heat Calculation

The specific heat, particularly at constant volume, is a thermodynamic quantity that measures how much energy is required to raise the temperature of a system. In statistical mechanics, it can be calculated by taking the derivative of the internal energy with respect to temperature. This involves differentiating the partition function or directly differentiating the expression for the internal energy, thereby linking microscopic energy states to macroscopic thermal properties.

Transcript

00:01 In this question, we have a rigid rotor.

00:05 So this rigid rotor is made up of a messless rod with two masses.

00:13 And 1 and 2.

00:17 And we are given that we are given that the moment of inertia of the situation about the center of mass is m1.

00:37 M2 divided by m1 plus m2 a square.

00:43 So this m1, m2 divided by m1 plus m2 is the reduced mass, which i'm going to call it mew, okay, a square, mew is the reduced mass.

00:57 Yeah.

01:01 So in part a, there are four parts in this question.

01:05 So basically we want to find the energy eigenvalues.

01:10 Function and the energy spectrum of the situation and then we use this idea to estimate the one length of carbon monoxide okay okay so in part a you want to show that the allow energies of the energy of the rigid rotor okay so um because for the first thing i will show kind of derive why the moments of inertia is that okay so if we take origin at m1 right and then the central mass will be m2 a divide by m1 plus m2 right so could be somewhere here is the center of mass okay and then this is m2a divided by m1 plus and then here would just be a minus m2 a by m1 plus m2 okay right so then the i of the center of mass the i about the center mess because the rotation is about the center of mass okay will be m1 times the distance square so this is m2 a divide by m1 plus m2 square and plus m2 and 2 a minus m2 a divide by m2 square okay so you get m1 and 2 square a square plus m2 so here when you combine this you get m1 a over m1 plus m2 so you have m2 times m1 square a square okay divide m1 plus m2 square okay so we put are the common factor which is m1 m2 a square and then you have m1 plus m2 right by m1 plus n2 square and you can see the common term and hence it gets the required expression and as mentioned i'm going to call this thing i'm going to call this term mu hey yeah the reduce less okay and then um classically when it comes to such rigid rotor, the kinetic energy, or the total energy is just half i omega square, then you can write it in terms of l square over i, okay, just like p square with 2m, then you have l square over 2i.

04:26 So the hematonian, h is l square over 2i, and then in quantum mechanics we know the eigenvalues of l square operator which is l, l plus 1, h bar square, right? but since we tend to use n in energy, using n as a quantum number, quantum number for energy.

05:15 Then the n is hbar square over 2 i n times n plus 1 okay and n is 0 1 2 and so on okay yeah shown okay so this is for part a then we want to part b i want to find a normalized eigen functions okay so the normalized my functions, okay, the other spherical harmonics, okay, because they are the spherical harmonics are the coordinate representation of l square operator.

06:07 Okay, so the thigh and m data phi is our ym.

06:14 M case so basically is just replacing the l in the y lm with n okay and you get the surica that you get the normalized eigen functions for this situation okay and we know that from the idea of angular momentum and will be an integer between minus n to n so n is minus n plus 1 in steps of 1 until you get n minus 1 and m okay so over here you have 2n plus 1 okay so the degeneracy is 2 n plus 1 okay so this is the answer okay for 2 for part b okay okay, next in practice you want to find a spectrum, the energy spectrum of this system.

07:24 Okay, so the frequency that is emitted from the emission spectrum will be the change in energy, the energy transition between levels, okay? so suppose we have transition from n plus one level to the end level.

08:16 Okay, so our delta e is equal to h -par square over 2i.

08:22 Okay, so we call that en is h -path -square over 2i.

08:27 2i n times n plus 1.

08:30 So the delta e would be from n plus 1 to n this is what i would do.

08:39 Yeah, so the first term is when you substitute n plus 1 into the n times n plus 1 and then you subtract n times n plus 1.

08:48 And so we have h plus square over 2 i 2 times n plus 1.

08:57 So the n plus 1 is common in the two terms and then you are left with n plus 2 minus n and then you are left with 2.

09:06 Okay and you can write this as h bar square over i times n plus 1.

09:13 So the two cancers out...