r₀ = r₁ + r₂.................................(1)
The molecules rotating end-over-end about a point "C" the centre of gravity. Then we have,
m₁ r₁ = m₂ r₂......................................(2)
The moment of inertia about "C" is defined as by,
I= m₁ r₁ ² + m₂ r₂²..............................(3)
l= m₁ r₁ r₁ + m₂ r₂ r₁
l= r₁ r₂ (m₁ + m₂)...............................(4)
m₁r₁ = m₂r₂ = m₂(r₀−r₁) = m₂r₀− m₂r₁
m₁r₁+ m₂r₁ = m₂r₀ (m₁+m₂)r₁= m₂r₀
r₁=m₂r₀/ (m₁ + m₂).............................(5)
m₂r₂ = m₁r₁ = m₁(r₀−r₂) = m₁r₀ −m₁r₂
m₂r₂ + m₁r₂ = m₂r₀
(m₁+ m₂) r₂ = m₂r₀
r₂ = m₂/ (m₁ + m₁)..............................(6)
Substitute the value of r₁ and r₂ from equation (5) and (6) into equation (4)
l = [m₁m₂/(m₁+m₂)]r₀² = μr₀²..............(7)
where μ = m₁m₂/(m₁+m₂) is called as reduced mass. Equation (7) defines moment of inertia in terms of the atomic masses and bond length. The expression for reduced mass can be written using the atomic weights M₁ and M₂ as
By solving Schrodinger equation for a rigid rotor, the energy Eⱼ for jth rotational level is obtained,
Eⱼ = [ h²/8π²l ] J (J+1) joules where J= 0,1,2,3,............(8)
h- planck's constant, l- moment of inertia either lB or l꜀ since both are equal. The term J can take integral values from zero, and is called the rotational quantum number. Equation (8) express the allowed rotational energies in joules. In the rotational spectra usually wave number units are used, so it useful to consider energies expressed in these units.
∈ⱼ= Eⱼ/hc = (h/8π²lc) J(J+1) cm²
Where J=0,1,2,3,4............................(9)
Where c is the velocity of light in cm S⁻¹since wave number is expressed in cm⁻¹
Equation (9) can be written as
∈j = BJ (J+1) cm⁻¹......................... (10)
"B" is the rotational constant is given by,
B= h/8π²lc cm⁻¹.............................. (11)
The allowed rotational energy values are calculated from equation (11). By substitution difference J values. Thus
∈₀= 0 cm⁻¹, ∈₁= 2B cm⁻¹ , ∈₂ = 6B cm⁻¹,
∈₃ = 8B cm⁻¹ etc. Below the diagram
Fig 1: The allowed rotational energy levels of a rigid diatomic molecules.
Fig 2: Allowed transition between the energy levels of the rigid diatomic molecule and the spectrum which arises from them.
Now consider the difference in energy between the levels ∆∈ = ∈ⱼ ₊₁ −∈ⱼ in cm⁻¹ units is given by
ʋ̄ = ∆∈ = ∈ⱼ₊₁− ∈ⱼ = BJ( J+1)(J+2) −BJ(J+1)
=B [J²+3J+2−J²−J] = 2B (J+1) cm⁻¹......(12)
The selection rule is ∆J = ± 1 i.e transition in which J changes one unit allowed. Other transitions are forbidden. Hence the equation (12) gives the rotational spectra for a rigid diatomic molecule which is shown in the above fig.2. It contains equally spaced lines with spacing equal to 2B cm⁻¹· Hence line separation 2B is determined from the rotational spectrum from which the moment of inertia and hence the intermolecular distance can be calculated.
Only if the diatomic molecule is heteronuclear, microwave spectrum will be observed since homonuclear diatomic molecule will have no dipolemoment change during the rotation and hence no interaction with radiation. Thus molecule such as HCl, CO will show rotational spectrum while N₂ and O₂ will not. But similarly the rotation about the bond axis is not microwave active. There are two reasons for this. Firstly the moment of inertia is very small about the bond axis. According to the equation (9) and (10), the energy levels are extremely widely spaced i.e. a molecules needs very high energy to undergo transition from J=0 → J+1 state which doesn't occur under normal spectroscopic conditions. Thus diatomic and all linear molecule are in the J=0 state for rotation about the axis and they are said to be not rotation. Such transition occurs there will be no dipole moment changes and hence no spectrum.
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