2. Heteronuclear Diatomic Molecule a. Complete the derivation of the moment of i

Question

2. Heteronuclear Diatomic Molecule a. Complete the derivation of the moment of inertia of the rigdly rotating diatomic molecule begun in class, and show that the monnent of inertia about the center of mass is 1 ! where is the reduced mass of the molecule. and r is the length of the molecule. Hint: use the fact that that the distances of either atom to the center of mass are related by mATA mra, and that b. Evaluate the ground and first excited rotational state energy levels of the CO molecule, in eV. You may take the masses of C and O to be 12 amu and 16 au respectively What is the energy of the transition from the first excited rotational state of CO to the ground state, in eV What part of the electromagnetic spectrum does this transition lie?

Explanation / Answer

2. a. consider a rigid body consisting of two bodies of masses Ma, Mb

then

distance of the center of mass from mass Ma be 'a'

and total seperation between the masses be r

then distance of Mb from center of mass is r - a

hence

moment of inertia of the two particles about center of mass = Ic

Ic = Ma*a^2 + Mb*(r - a)^2

now,

from the definition of center of mass

Ma*a = Mb*(r - a)

hence

Ma*a = Mb*r - Mb*a

a = Mb*r/(Ma + Mb)

hence

Ic = Ma*Mb^2*r^2/(Ma + Mb)^2 + Mb(r - Mb*r/(Ma + Mb))^2

Ic = Ma*Mb^2*r^2/(Ma + Mb)^2 + Mb*r^2(Ma + Mb - Mb)^2/(Ma + Mb)^2

Ic = Ma*Mb^2*r^2/(Ma + Mb)^2 + Mb*r^2(Ma)^2/(Ma + Mb)^2

Ic = Mb*Ma(Mb + Ma)*r^2/(Ma + Mb)^2

Ic = Mb*Ma*r^2/(Ma + Mb) = mu*r^2

where mu = Mb*Ma/(Ma + Mb)

1/mu = 1/Ma + 1/Mb

b. now rotaitonal energy levels are given by

Ej = BJ(J + 1)

for ground state, J = 0

Ej = 0

for first excited atate

J = 1

Ej = 2B = h^2/4*pi^2*mu*r^2

now, for CO, r = 116 pm

Ma = 12 amu

Mb = 16 amu

hence

Ej = 4.770817*10^-4 eV

c. energy of transition = Ej - 0 = Ej = 4.770817*10^-4 eV

this coresponds to frequency f

hf = Ej

f = 1.1444239 *10^11 Hz

this corresponds to microwave part of the spectrum

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