Question
Asked Nov 10, 2019
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Calculate the frequency of the J = 7 ← 6 transition in the pure rotational spectrum of 12C16O. Assume the equilibrium bond length is 109.36pm.

Frequency is _____ H

What is the corresponding wavenumber? ______ cm-1
 
 
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Expert Answer

Step 1

The energy in the term of corresponding wavenumber for the rotational region spectra (ƐJ) is calculated as show in equation (1) where h and c is the Planck’s constant and speed of light respectively, I is the moment of inertia, and J is the rotational energy level.

h
J (J+1 )
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h J (J+1 )

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Step 2

The emitted energy in the term of wavenumber is calculated as shown in equation (2). This will give the equation (3).

h
[7(7+1)- 6(6+1)]..2)
EJ-J6)
h
[56-42]
h
(14).(3)
EJ-HJ6) 8T2Ic
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h [7(7+1)- 6(6+1)]..2) EJ-J6) h [56-42] h (14).(3) EJ-HJ6) 8T2Ic

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Step 3

The moment of inertia for the given molecule is calculated as shown in equation (4) where MC12 and MO16 are molar masses of carbon and oxygen, and r is the bond length. The values for r, MC12, and MO16 are substituted...

М М,
c12
1
o16
I =
g molx
-- (4)
- xp2
M2M
6.022 x 1023 mol
12x16
1
g molx
- x (109.36x 10-12 m
12 16
6.022 x 1023 mol
g m2x10 kg/g
= 1.362x 103
= 1.362x106
kg m2
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М М, c12 1 o16 I = g molx -- (4) - xp2 M2M 6.022 x 1023 mol 12x16 1 g molx - x (109.36x 10-12 m 12 16 6.022 x 1023 mol g m2x10 kg/g = 1.362x 103 = 1.362x106 kg m2

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