Rotational spectrum of diatomic molecules-

In this particular article Rotational spectrum of diatomic molecules, we are going to discuss the diatomic molecule as a rigid rotator along with its eigenvalue and eigenfunction.

The diatomic molecule as a Rigid rotator-

Let us consider HCl, a linear molecule as an example. Let us consider that two atoms having mass m₁ and m₂ are joined together by a line, whose length is-

r = r₁+r₂

Again let that bond length of the molecule doesn’t change during rotation. We also assume that the atoms are point-like. A diatomic molecule whose nuclei are separated by a finite distance behaves as a rigid- rotator.

m₁r₁ = m₂r₂

The moment of inertia of the molecule-
I = m₁r₁²+ m₂r₂²
We know that-
r= r₁+r₂
So, r ₁ = m₂r/m₁+m₂
r₂ = m₁r /m₁+m₂
Putting in I,
I= m₁ (m₂r/m₁+m₂²)² + m₂(m₁r/m₁ + m₂)²
By solving this equation we get-
I = (m₁.m₂/m₁+m₂)r²
I= μr²
Where μ is called reduced mass.
And μ = m₁m₂/m₁+m₂

The above equation defines the moment of inertia of the diatomic molecule in terms of the atomic masses and the bond length. Hence, we can say that the moment of inertia of the diatomic molecule shows resemblance with that of a mass point of mass μ at a distance R from the axis.


Wave Equation for a rigid rotator-

The kinetic energy of a particle of mass m in the Cartesian coordinate system is-
T= m/2 [( dx/dt)²+ (dy/dt)²+(dz/dt)²]
(Note:- we know that the kinetic energy of a particle is given by
E= mr²/2)

Again, in the spherical cartesian coordinate system-
T= m/2 [( dr/dt)²+r²(dθ/dt)²+r²sin²(dφ/dt)²]
If the distance of the particle from the origin is constant then,
T = mr²/2 [( dθ/dt)²+ sin²θ (dφ/dt)²]                 (1)

For a diatomic particle system-
T= 1/2 (m₁r₁² + m₂r₂²) [ (dθ/dt)²+ sin ²θ(dφ/dt)²]

The total energy of the rotator,

E= T+V

Where T shows kinetic energy and V shows potential energy.

E= I/2 [(dθ/dt)²+sin²θ(dφ/dt)²]+V

For the free particle, we know that V=0

E= I/2 [(dθ/dt)²+sin²θ(dφ/dt)²]                     (2)

In this system, free rigid rotator behaves as a free single particle and the mass of this rotator is m=I when r=1. Then Schrodinger equation in the spherical coordinate system,

[ 1/r²(δ/δr)(r²δ/δr) + 1/r²sinθ (δ/δθ) (sinθδ/δθ) + 1/r²sin²θ(δ²/δφ²)]ψ+(2m/h²)(E-V)ψ]=0

(Note- where h is Planck’s constant)

For rigid rotator,

m=I, r=1 and V=0

Putting these values,

[  sinθ (δ/δθ) (sin(θδ/δθ)) + sin²θ(δ²/δφ²)]ψ+(2m/h²)(Er)ψ]=0

The solution of the wave equation for rigid rotator-

By the method of separation of variables, we know that-

ψ(θ,φ) = Θ(θ)Φ(φ)

Put this value in the above equation and multiply by sin²θ/Θ(θ)Φ(φ), we get

(sinθ/θ)d/dθ (sinθ δΘ(θ)/δθ)+ 2I/h²(Er sin²θ) = −1/Φ(φ)(δ²Φ(φ)/δφ²)

Let LHS = m²

And, RHS= m²

Here m² is constant.

(sinθ/θ)d/dθ (sinθ δΘ(θ)/δθ)+ 2I/h²(Er sin²θ) =m²

(sinθ/θ)d/dθ (sinθ δΘ(θ)/δθ)+ (2IEr/h²)sin²θ) =m²

Put ( 2IEr/h²) = β

Or , (sinθ/θ)d/dθ (sinθ δΘ(θ)/δθ)+ (β)−m²/sin²θ)Θ(θ) =m²     (a)


−1/Φ(φ)(δ²Φ(φ)/δφ²) = m²

(δ²Φ(φ)/δφ²) + m²Φ(φ)=0

The solution of this equation is given by

Φ(φ)=A exp(imφ)

Where A is a normalization constant and m can have positive as well as negative values. Since values of φ differing by integral multiples of 2π refer to the same point and Φ(φ) has to be e single-valued function of φ.

exp (i2mπ) = 1

This restricts m to a discrete set

m=0,±1 ±2 ±3…

The normalization condition is obtained from the normalization condition :

∫ φ*mφn dφ = 1 it is

A= 1/√2π

The solution of Equation (a)

To find the solution of equation (a), it is convenient to use substitution x= cosθ

Or in other words, 1- x²=sin²θ

Hence above equation i.e equation (a) can be written as –

δ/δx { (1-x²)dθ/dx} + { β−m²/1−x²} Θ(θ) =0

This is called Legendre’s equation.

When θ varies from 0 to π, then x varies from +1 to -1.

According to Quantum mechanics, ψ should be single-valued, finite and continuous. And above equation gives solution only when β = J(J+1)

Where J = 0,1,2,3,…. rotational quantum number.

And the solution of  Legendre’s equation is given by-

The eygen function of the rigid rotator

We know,

ψ(θ,φ) = Θ(θ)Φ(φ)

Put the solution

Now applying normalisation condition

N= 1/√2π [ (2J+1)(J- |m|)!/ 2(J + |m|)!] ¹/₂

N= 1/√2π [ (2J+1)(J- |m|)!/ 2(J + |m|)!] ¹/₂ exp (imφ) pmj (cosθ)

i.e. eygen function of rigid rotator

Eygen value of rigid rotator

We have , β = J(J+1) and

β = 2IEr/h²

Hence, comparing both equation we get-

J(J+1) = 2IEr/h²

And, Er= J(J+1)h²/2I

This equation shows that energy levels are quantized for rigid rotator.





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