The nucleus ^2 H has no rotational excited states. Justify this observation with a calculation

step 1 In this problem, we're looking at a diatomic molecule. We're contrasting the inertia and the ene

The nucleus ^2 H has no rotational excited states. Justify...

Key Concepts

Rigid Rotor Model

This concept describes a system in which the body rotates about a fixed axis without internal deformation. In quantum mechanics, it is used to derive the energy levels of rotating systems by applying the quantization of angular momentum, yielding discrete rotational energy states based on the moment of inertia of the rotating body.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. In the context of the nucleus, it is calculated based on the mass distribution and size of the nucleus. Since nuclear radii are extremely small, the resulting moment of inertia is also very small, leading to very large spacings between theoretical rotational energy levels.

Quantization of Angular Momentum

In quantum systems, angular momentum is quantized according to the relationship J(J + 1), where J is the angular momentum quantum number. This quantization directly affects the rotational energy levels of a system, as these levels are given by the formula E = (?²/2I)J(J + 1), meaning that larger energy gaps can arise if the moment of inertia (I) is small.

Nuclear Size and Energy Scale

Nuclei are very compact systems, and when estimating the energy of rotational excitations, the small size (and hence small moment of inertia) implies that any rotational excited state would require an enormous amount of energy, often exceeding typical nuclear excitation energies. This explains why some nuclei, like the deuteron, do not exhibit observable rotational excited states, as the required transition energies lie far beyond accessible experimental ranges.

Transcript

00:01 In this problem, we're looking at a diatomic molecule.

00:03 We're contrasting the inertia and the energy between the cases of rotation about different axes.

00:16 So in one case, we have the ordinary familiar situation up to now.

00:22 The axis perpendicular to the bond.

00:27 So that's this x -axis.

00:29 It's a rotation along the x -axis, which is perpendicular to the bond.

00:34 That is the reduced mass times the radial distance between the center of mass and the two -point masses.

00:46 But there's another possibility, which the problem is focusing on, that usually gets ignored.

00:52 That's when the rotation is along the same axis as the bond length.

00:58 So basically the molecule, the bond like isn't moving in space.

01:06 The balls, these spheres are just spinning basically in place.

01:12 That's the kind of rotation it's talking about.

01:16 And the point of the problem is that that kind of rotation occurs in a much higher energy level to be excited into.

01:25 So that's why it's usually doesn't get ignored.

01:28 So to calculate this problem, the difference is that in that case, you're talking about spinning spheres, solid spheres, where it's modeling it as a uniform density of solid matter of a certain mass.

01:42 They have equal and fixed radius of two femtometers, 10 to negative 15 meters.

01:50 And the inertia equation for solid spheres like that is two fifths times mass times the radius of the sphere.

01:59 But whereas for the radius between the two masses along the y -axis, that radius in capital r is two times 10 to negative 10 meters, so two angstroms for each.

02:19 And the masses are equal.

02:25 So what we're looking at here is the energy, the rotational energy comes from dividing the energy.

02:35 Term and that has a radius squared element in the inertia.

02:45 So there's a five order of magnitude difference between the radius of the spheres and the radius between the origin and the masses.

02:54 So when you square it does a 10 order magnitude difference.

02:58 That leads to the 10 order magnitude difference between the energies of those two cases.

03:07 So we're going to demonstrate that.

03:10 The reduced mass situation.

03:15 Since the masses are equal, that becomes m squared divided by 2m.

03:23 So mu is just m divided by 2.

03:30 So let's see.

03:34 So first we want to find the inertia is about the x and y axes.

03:40 So we use these equations, right? so it's just, well, for the rotation about the x -axis, we just use this reduced mass equation.

04:01 But for the other one, we'll individually add these terms for the two spheres.

04:08 So i'll do the sphere case first.

04:12 So the inertia, rotation by the y -axis, that becomes.

04:17 Comes two -fifths, well, i'll write it generally first, plus two -fifths into r2 squared.

04:40 But the r -1s and the r's and the atoms are the same.

04:48 It's like an identical clone situation...

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