{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10 Lecture19 Feb 22 - CHEM 350 Lecture 19 • Rotations...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM 350 Lecture 19, February 22 2010 • Rotations . First let us review briefly the quantum mechanical description of simple rotational motion for a rigid rotor: ~ 2 c J 2 2 μR 2 e ψ ( θ,φ ) = E rot ψ ( θ,φ ) , Upon setting μR 2 e = I (the moment of inertia of the rotor), we obtain E rot = ~ 2 2 I j ( j + 1) = j , j = 0 , 1 , 2 , ··· The ψ j,m j ( θ,φ ) = Y j,m j ( θ,φ ) are the spherical harmonics: for each value of j , there will be 2 j + 1 values of m j and, as the energy depends only upon j , each level will be (2 j + 1)–fold degenerate! q rot ( T ) ≡ X states e- βE state = X j,m j e- β j,m j ≡ X levels Ω level e- β level = ∞ X j =0 (2 j + 1)e- β j rotational temperature: Θ rot = ~ 2 2 Ik B from βE rot = β ~ 2 2 I j ( j + 1) ≡ Θ rot T j ( j + 1) (i) For temperatures T such that T Θ rot , q rot ( T ) can be written as q rot ( T ) ’ 1 + 3e- 2Θ rot /T + 5e- 6Θ rot /T + ··· (ii) For temperatures T such that T Θ rot , we have q rot ( T ) = ∞...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

10 Lecture19 Feb 22 - CHEM 350 Lecture 19 • Rotations...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online