Quantum chemistry

1. THE RIGID ROTATOR
Name: D.Muthu pandi
Roll no: 22APCH017
Class : I- M.sc Chemistry

2.  The rotation of diatomic molecule in Space where the bond length is assumed to remain unchanged
during rotation
 The theory of Such rigid rotator Space is useful in dealing with rotational spectra of diatomic molecules.
 This the basic Principle of rotational spectroscopy. It occurs in Microwave region of the series.
Rigid Rotator:
 Let us Consider homonuclear diatomic molecule There are 3 types of rotation. Along x, y and z axis,
But the internuclear axis of rotation is negligible
 It can rotate about. an axis is perpendicular to the internuclear
distance (i.e. y and z axis)

3. The Postulate of Quantum mechanics,
ĤΨ = 𝐸Ψ …….(1)
Ĥ = KÊ+𝑃E
This case(rigid rotator motion) potential energy is zero, because no force of attraction.
Why this called rigid rotator,
The internuclear distance does not change so they called rigid rotator.
K.E=
1
2
𝑚𝑣2
In linear momentum
K.E=
1
2
𝐼ω2
Angular momentum (m=I)
I = mr2
μ=
𝑚1
𝑚2
𝑚1
+𝑚2
I is moment of inertia
In this case m is reduce mass
KÊ =
𝐼(𝐼ω2
)
2𝐼
=
𝐿2
2𝐼
= Ĥ (K.E=
𝑃2
2𝑚
)

4. The Schrodinger wave equation, can not be Solved readily Cartesian equation So we use for spherical Polar
Coordinates, Instant of Cartesian Polar Coordinate
(r,θ,Φ spherical polar coordinates)
r= 0 to ∞
Θ= 0 to 2
Φ= 0 to
𝐿2= ħ2 [
1
sin θ
𝜕
𝜕𝜃
(sin 𝜃
𝜕
𝜕𝑥
) +
1
sin2
𝜃
𝜕2
𝜕2
∅
] ……(2)
ħ2
2𝐼
[
1
sin θ
𝜕
𝜕𝜃
(sin 𝜃
𝜕Ψ
𝜕𝑥
) +
1
sin2
𝜃
𝜕2
Ψ
𝜕2
Φ
]= EΨ ……(3)
The total wave function
Ψ(𝜃, ∅)= θ 𝜃
Φ(∅)
Total spherical harmonics
Separation of variables:
(L.H.S)
sin 𝜃
θ
𝜕
𝜕𝜃
(sin 𝜃
𝜕θ
𝜕𝑥
) +
8𝜋2
𝐼𝐸
ħ2 sin2 𝜃

5. = -
1
Φ
𝜕2
Φ
𝜕2
Φ
(R.H.S)
L.H.S= R.H.S = m2 m2= constant
β=
8𝜋2
𝐼𝐸
ħ2
-
1
Φ
𝑑2
Φ
𝑑2
Φ
= m2
𝑑2
Φ
𝑑2
Φ
+ m2Φ= 0
Φ(∅) =Ne
± im∅ ‘N’ is normalization constant
The wave function is available of(Φ) is acceptable when m= 0,±1,±2,…. ∅= 0 to 2
by applying normalization condition the value of ‘N’ can be obtained
Φ(∅) =
1
2𝜋
e
± im∅ (Normalized function)
L.H.S.
On solving LHS equation (depends on only θ) using assoiated legendre polynomial
(PL
𝑚
cos 𝜃 )

6. θ(𝜃)= N PL
𝑚
cos 𝜃
N = (
2𝑙+1
2
×
(𝑙− 𝑚 !
(𝑙+ 𝑚 !
)
1
2
β = l (l+1) ‘l’ is rotational quantum number
The energy of eigen values are,
β=
8𝜋2
𝐼𝐸
ħ2 = l (l+1)
E =
𝑙 𝑙+1 ℎ2
8𝜋2
𝐼
(l=0,1,2,…..)
In rotational(mw) spectroscopy,
E =
τ τ+1 h2
8π2
I
(𝜏 =0,1,2,…….)
I (𝜇r2) 𝜇 =
𝑚1
𝑚2
𝑚1
+𝑚2
From RS bond distance is obtained

7. THANK YOU!