ROTATIONAL (MICROWAVE) SPECTRA OF DIATOMIC MOLECULES

Consider a diatomic molecule in which m₁ and m₂ are the masses of the two atoms joined by a rigid bar (the bond) whose length is-----------

                      r = r₁ + r₂    ----------------------   (1)

Rotating about an axis passing through its centre of gravity.

The centre of gravity is defined by the equality of the moments about it, that is ----------

m₁r₁ = m₂r₂   -------------------- (2)

                                                                              

The moment of inertia I of a molecule is defined as ---

 I = m₁r₁² + m₂r₂²

   = m₂r₂r₁ + m₁r₁r₂ (From Eq 2) -------(3) 

   = r₁r₂ (m₁ + m₂)

From equation (1) and (2) we get --------

m₁r₁ = m₂r₂ = m₂(r-r₁)

Therefore,

r₁ = (m₂r)/(m₁+m₂)  and r₂ = (m₁r)/( m₁+m₂) ---- (4)

Replacing (4) in (3) we get,--------------

I = [(m₁m₂)/(m₁+m₂)]r²   = µr² ------- (5)

where, µ = [(m₁m₂)/(m₁+m₂)], and µ is called the reduced mass of the molecule.

By the use of the Schrodinger equation it may be shown that the rotational energy levels allowed to the rigid diatomic molecule are given by the expression---

E_J = [h²/8π²I]J(J+1) joules  where J = 0,1,2-----   (6)

Where h is Planck’s constant, and I is moment of inertia and J is rotational quantum number.

We can write the equation (6) as -----

ε_J = E_J/hc = [h/8π²Ic]J(J+1)cm^-1(J = 0,1,2--) (7)

Where c is the velocity of light expressed in cm/sec.

We can write the equation (7) as---

ε_J = BJ(J+1) cm^-1  (J = 0,1,2,----) 

Where B , the rotational constant, is given by -----

B = [h/8π²Ic] cm^-1    

The rotational transitions for a rigid diatomic molecule are governed by the selection rule-----

ΔJ = ± 1

That is only those transition are allowed in which the rotational quantum number changes by unity. The + sign refers to absorption and the – sign to emission of radiation.

For a transition taking place from J to J+1, the rotational frequency is given by----- 

ν_(J→J+1) = B(J+1)(J+2)- BJ(J+1)

              = B(J² +3J+2)- B(J² +J)

              = 2B(J+1) cm^-1    

Thus, ---------------

ν_(0→1) = 2B

ν_(1→2) = 4B

ν_(2→3) = 6B

ν_(3→4) = 8B

We see that rotational spectrum of a rigid diatomic molecule consists of a series of lines at 2B, 4B, 6B, 8B etc.These lines are spaced by an amount of 2B.

    

                 

                

Rotational spectrum of a rigid diatomic molecule.