Quantum Chemistry/Example 15

Write an example question showing the determination of the bond length of CO using microwave spectroscopy

Example 15:

Part A:

The exact atomic mass of ${\displaystyle ^{12}C}$  is 12.011 amu and ${\displaystyle ^{16}O}$  is 15.9994 amu [1]. Solve for the reduced mass of ${\displaystyle ^{12}C^{16}O}$  in units of kg.

Solution:

${\displaystyle \mu =\left({\frac {m_{1}m_{2}}{m_{1}+m_{2}}}\right)}$ [2]

${\displaystyle \mu =\left({\frac {(12.011amu)(15.9994amu)}{12.011amu+15.9994amu}}\right)}$

${\displaystyle \mu =6.860622961amu}$

${\displaystyle \mu =(6.860622961amu)(1.67377\times 10^{-27}\left({\frac {kg}{amu}}\right))}$

${\displaystyle \mu =1.140\times 10^{-26}kg}$

Part B:

In microwave spectroscopy, the formula of the rotational constant is given below. Rearrange this formula to solve for the bond length.

${\displaystyle \beta =\left({\frac {h}{8\pi ^{2}c\mu R^{2}}}\right)}$ [3]

Solution:

${\displaystyle \beta =\left({\frac {h}{8\pi ^{2}c\mu R^{2}}}\right)}$

${\displaystyle R^{2}\cdot \beta =\left({\frac {h}{8\pi ^{2}c\mu }}\right)}$

${\displaystyle R^{2}=\left({\frac {h}{8\pi ^{2}c\mu \beta }}\right)}$

${\displaystyle R={\sqrt {\left({\frac {h}{8\pi ^{2}c\mu \beta }}\right)}}}$

Part C:

Using the rearranged formula from part B, determine the bond length of ${\displaystyle ^{12}C^{16}O}$  in units of angstrom (A) given that the rotational constant ${\displaystyle (\beta )}$  for ${\displaystyle ^{12}C^{16}O}$  is ${\displaystyle 1.93128087cm^{-1}}$ [4] .

Solution:

${\displaystyle R={\sqrt {\left({\frac {h}{8\pi ^{2}c\mu \beta }}\right)}}}$

${\displaystyle R={\sqrt {\left({\frac {6.6260755\times 10^{-34}kg\cdot m^{2}}{8\pi ^{2}(2.99792458\times 10^{10}cm/s)(1.139233885\times 10^{-26}kg)(1.93128087cm^{-1})}}\right)}}}$

${\displaystyle R={\sqrt {1.27229439\times 10^{-20}m}}}$

${\displaystyle R={\sqrt {1.27229439\times 10^{-20}m}}\times \left({\frac {10^{10}\mathrm {A} }{m}}\right)}$

${\displaystyle R=1.128\times 10^{-10}\mathrm {A} }$