The document discusses applications of microwave spectroscopy, focusing on determining the moment of inertia and bond length of diatomic molecules using rotational spectra. It explains the isotope effect and provides formulas and numerical examples for calculations. Students are advised on the correct methods to approach problems related to microwave spectroscopy.

1. Dr. Y. S. THAKARE
M.Sc. (CHE) Ph D, NET, SET
Assistant Professor in Chemistry,
Shri Shivaji Science College, Amravati
Email: yogitathakare_2007@rediffmail.com
B Sc- III Year
SEM-V
PAPER-III
PHYSICAL CHEMISTRY
UNIT- VI
Applications of Microwave Spectroscopy
02-November -20 1

2. We know that
I = µ𝒓𝒐
𝟐
µ =
𝒎𝟏𝒎𝟐
𝒎𝟏+𝒎𝟐
𝐁 =
𝒉
𝟖𝝅𝟐𝑰𝑪
𝑪𝒎−𝟏
Mass is more…..
Reduced mass is more….
Moment of inertia (I) is more….
Rotational Constant (B) is less…..

3. Isotope Effect
If one of the atoms in a heteronuclear diatomic molecule (AB) is
substituted by the isotope of the same element with higher mass
number (A’B) then the position of the absorption line in a rotational
spectra of these two molecules will be different. This phenomenon
is called as isotope effect.
For example:- HCl35 and HCl37
Isotopically substituted molecule (A’B) will have higher reduced
mass than AB. Since the bond length (r0) is same for both, A’B will
have higher moment of inertia than AB. Rotational constant is
inversely proportional to the moment of inertia and hence
rotational constant of A’B will be lower than that of AB. Rotational
spectra of these two molecule are shown in following figure.
Isotope effect is very useful in the calculation of isotopic masses.

5. Applications of Microwave Spectroscopy
1) Determination of moment of inertia (I) of the diatomic
molecule from the recorded absorption rotational spectrum, the
spacing between the consecutive lines (2B) is calculated. From this I
is calculated using the formula
I =
h
8π2BC
Where, h = 6.62 × 10-34 Js
𝜋 = 3.14
B =in m-1 (1 cm -1 = 100 m-1)
C= 3 × 108 ms-1
I = in kg m2

6. 2) Determination of bond length (r0) for diatomic molecules
First the moment of inertia (I) for a given molecule is calculated from
the recorded spectrum as stated above. Then the reduced mass (µ)
of the molecule is calculated from atomic masses using the formula
μ =
m1xm2
m1+m2
in kg
Finally , the bond length is calculated by using the formula
I = µr0
2
r0
2 =
I
μ
r0=
I
μ
Since the bond lengths, are small distances, these are usually
expressed in nm (1 nm = 109m) or less frequently in A0 (1 A0 = 10-10
m)

7. Numericals on Microwave Spectroscopy
While solving Numericals on microwave spectroscopy, given
data should be transformed in SI units (to avoid confusion) and all
the calculation should be carried out in SI system of units. Students
should remember the value of following constants and use these
values in calculations. (if these are not provided in questions paper )
Planck`s constant, h = 6.62 × 10-34 Js
Velocity of light, C= 3 × 108 ms-1
= 3 × 1010 cms-1
π = 3.14
1 amu = 1.66 × 10-27 kg
1
𝑁𝐴
=
1
6.023 × 10−23
= 0.166 × 10−23
𝑔
= 0.166 × 10−26𝑘𝑔
= 1.66 × 10−27𝑘𝑔

8. For developing confidence, students should proceed as follows
1. Read the numerical very carefully with concentration and note
down the given information.
2. Write parameter or variable to be evaluated
3. Write the appropriate formula correctly.
4. Substitute the values with proper units.
5. Workout the required arithmetic calculations
(additions / subtraction/ multiplication/ division)
6. Write the answer with proper units.

9. Example 1 : Internuclear distance in HF molecules is 0.092nm.
Calculate moment of inertia of HF molecule.
mH = 1.00 amu, mF = 19.00 amu (1 amu = 1.661 x 10-27kg)
Solution:-
Give that
𝑟0 = 0.092𝑛𝑚 = 0.092 × 10−9𝑚
𝑚𝐻 = 1.00𝑎𝑚𝑢
𝑚𝐹 = 19.00𝑎𝑚𝑢
1 𝑎𝑚𝑢 = 1.661 × 10−27𝑘𝑔
𝐼 = ?
We know that
I = µ𝒓𝒐
𝟐
μ =
m1x m2
m1 + m2
=
𝑚𝐻 × 𝑚𝐹
𝑚𝐻 + 𝑚𝐹
=
1 × 19
1 + 19
= 0.95𝑎𝑚𝑢 = 0.95 × 1.661 × 10−27𝑘𝑔
𝐼 = 𝜇 𝒓𝒐
𝟐 = 𝟎. 𝟗𝟓 × 1.661 × 10−27𝑘𝑔 × (0.092 × 10−9𝑚)2
𝐼 = 1.33x10-47 kgm2

10. Example 2 : The rotational spectrum of HF has a lines 4190 m-1 apart.
Calculate moment of inertia and bond length in HF. Given mH = 1amu,
mF = 19 amu, C = 2.998 x 108m/s, h = 6.626 x 10-34Js
Solution : 2B = 4190 m-1 B = 2095 m-1
I =
h
8π2B.C
=
6.626x 10−34
8x (3.14)2 x 2095 x 3 x 108 = 1.3374 × 10−4𝑘𝑔𝑚2
Also,
μ =
m1x m2
m1+ m2
=
𝑚𝐻×𝑚𝐹
𝑚𝐻+𝑚𝐹
=
1×19
1+19
= 0.95𝑎𝑚𝑢 = 0.95 × 1.661 × 10−27𝑘𝑔
= 1.578 × 10−27
𝑘𝑔
𝐼 = 𝜇 𝒓𝒐
𝟐
𝒓𝒐
𝟐=
𝐼
𝜇
𝑟0 =
𝐼
𝜇
=
1.3374 × 10−47𝑘𝑔𝑚2
1.578 × 10−27𝑘𝑔
= 0.92 𝑋 10−10𝑚
= 0.092 × 10−9𝑚 = 0.092 nm

11. Example 3 : The Rotational spectrum of CO shows a series
of equidistance lines having spaced 3.84235 cm-1apart. Calculate
moment of inertia and bond length of C0
C = 12 amu ; O = 16 amu
Solution : Given that
2B = 3.84235 cm-1 = 3.84235 x 102m-1 = 384.235 m-1
𝑚𝐶 = 12.00𝑎𝑚𝑢
𝑚𝑂 = 16.00𝑎𝑚𝑢
1 𝑎𝑚𝑢 = 1.661 × 10−27
𝑘𝑔
𝐼 = ?
𝑟0= ?
We know that 𝑰 =
h
8π2B.C
OR I =
h
4π2(2𝐵)C
=
6.626x 10−34
4x (3.14)2 x 384.235 x 3 x 108 = 0.0001457x 10−42 =14.57 x 10−47 kg
m2
μ =
m1x m2
m1+ m2
=
𝑚𝐶×𝑚𝑂
𝑚𝐶+𝑚𝑂
=
12×16
12+16
= 6.857𝑎𝑚𝑢 = 6.857 × 1.661 × 10−27𝑘𝑔
= 11.3810−27
𝑘𝑔
𝐼 = 𝜇 𝒓𝒐
𝟐 OR 𝒓𝒐
𝟐=
𝐼
𝜇
𝑟0 =
𝐼
𝜇
=
14.57 × 10−47𝑘𝑔𝑚2
11.3810 × 10−27𝑘𝑔
= 1.13 × 10−10
𝑚 = 0.113 × 10−9
𝑚 = 0.113 nm
1m=100cm=102 cm
1m-1 =10-2cm-1
cm-1 =102m-1

12. Example 4. For HCl, the frequency difference between successive absorption lines
has been found to be 20.7 cm-1. Find the bond length for HCl molecule in the
rotational spectrum. mH = 1.08 amu ; mCl = 35.5 amu
Solution : 2B = 20.7 cm-1 Or B = 10.35 cm-1= 1035 m-1
I =
h
8π2BC
=
6.62 x 10−34
8 x 3.14 2x 10.35 x 3 x 108 ×102 = 2.70x 10-47 Kg m2
Reduced mass of HCl molecule is given by
μ =
m1x m2
m1+ m2
=
𝑚𝐻×𝑚𝐶𝑙
𝑚𝐻+𝑚𝐶𝑙
=
1×35.5
1+35.5
= 1.04 𝑎𝑚𝑢 = 1.04 × 1.661 × 10−27𝑘𝑔
= 1.73 𝑥 10−27𝑘𝑔
𝑟0 =
𝐼
𝜇
=
2.70 𝑥10−47𝑘𝑔𝑚2
1.73 × 10−27𝑘𝑔
= 1.25 × 10−10𝑚 = 0.125 × 10−9𝑚
= 0.125 nm

13. Example 5. The microwave spectrum of HI molecule consists of a series of
equidistant lines with spacing of 12.8 cm-1.Calculate the bond length for HI. mH =
1 amu ; mI = 127 amu
Solution : 2B = 12.8 cm-1 ∴ B = 6.4cm-1= 6.4 × 102 m-1
I =
h
8π2B.C
=
6.62 x 10−34
8 x 3.14 2x 6.4 × 102 ×3 × 108
= 4.37 × 10-47 kgm2
µ =
1 ×127
(1+127)
= 0.992 amu
= 0.992 × 1.66 × 10-27 kg = 1.647 × 10-27 kg
r0
2
=
I
μ
=
4.37 x10−47
1.647 x 10−27= 2.65 × 10-20 m2
r0=1.63 × 10-10 m
= 0.163 nm