# 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 $^{12}C$  is 12.011 amu and $^{16}O$  is 15.9994 amu . Solve for the reduced mass of $^{12}C^{16}O$  in units of kg.

Solution:

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

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

$\mu =6.860622961amu$

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

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

Part B:

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

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

Solution:

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

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

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

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

Part C:

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

Solution:

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

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

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

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

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