a.) The spacings of the rotational fine structure lines of carbon dioxide 12C16O2 are determined from IR spectroscopy to be 0.7604 cm-1. Calculate the C=O bond length of the molecule (given masses: m(12C) = 12 amu, m(16O) = 15.9949 amu).Ans = ______ Angstromsb.) Suppose that the wavenumber of the J = 1 ← 0 rotational transition of 1H79Br considered as a rigid rotor was measured to be 17.89 cm-1, what is the bond length? Ans = ____  AngstromsWhen the moment of interia is 3.132x10^-47 kg*m^2(Given the isotopic masses:(m(79Br) = 78.9183 amu, m(81Br) = 80.9163 amu)

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Asked Nov 25, 2019
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a.) The spacings of the rotational fine structure lines of carbon dioxide 12C16O2 are determined from IR spectroscopy to be 0.7604 cm-1. Calculate the C=O bond length of the molecule (given masses: m(12C) = 12 amu, m(16O) = 15.9949 amu).
Ans = ______ Angstroms

b.) Suppose that the wavenumber of the J = 1 ← 0 rotational transition of 1H79Br considered as a rigid rotor was measured to be 17.89 cm-1, what is the bond length? Ans = ____  Angstroms

When the moment of interia is 3.132x10^-47 kg*m^2
(Given the isotopic masses:(m(79Br) = 78.9183 amu, m(81Br) = 80.9163 amu)

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Step 1

The bond length of a molecule in IR spectroscopy is calculated using the formula :

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8 п2 с В и Where, R bond length h Planck's constant = 6.626x10-34 kg m2 s1 c speed of light = 3.0x108 m s1 В wave number ureduced mass

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Step 2

Given,

Spacing of the rotational fine structure lines, 2B = 0.7604 cm-1

Atomic mass of C, m1 = 12 amu

Atomic mass of O, m2 = 15.9949 amu

The rotational spectra of a diatomic molecule have several peaks spaced by 2B.

Therefore, the value of B can be calculated as:

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2B 0.7604 cm 0.7604 cm 1 В - 2 В - 0.3802 cm-1 В - 38.02 m-1

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Step 3

The reduced mass can be...

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mim2 (m1 m2) NA 12 x 15.9949 (12 15.9949) x 6.022 x 1023 1.14x10-23 g 1.14x1026 kg

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