Question

The nuclei of the O2molecule are separated by 1.20 × 10–10m. The mass of each oxygen atom in the molecule is 2.66 × 10–26kg. (a) Determine the rotational energies of an oxygen molecule in electron volts for the levels corresponding to J = 0, 1, and 2.(b) The effective force constant k between the atoms in the oxygen molecule is 1177 N/m. Determine the vibrational energies (in electron volts) corresponding to v = 0, 1, and 2.

Answer 1

(a) The reduced mass of the O₂ is

$\mu= \frac{m_om_o}{m_o+m_o} \\ = \frac{16.00 \times 16.00}{16.00+16.00} = 8 \;u \\ = 8 \times 1.66 \times 10^{-27} = 1.33 \times 10^{-26} \; kg$

The momentum inertia is

$I = \mu r^2 \\ = 1.33 \times 10^{26} \times (1.20 \times 10^{-10})^2 \\ = 1.91 \times 10^{-46} \; kg \times m^2$

The rotational energies are

$E_{rot}= \frac{h^2}{2l}J(J+1) \\ = \frac{(6.626 \times 10^{-34 \;J \;s/2 \pi )^2}}{2 \times 1.91 \times 10^{-46} \;kg \; m^2}J(J+1) \\ = (2.91 \times 10^{-23}\;J) J(J+1)$

For J=0,1,2

$E_{rot}= 0, 3.63 \times 10^{-4}\;eV, 1.09 \times 10^{-3}\;eV$

(b) The vibrational energies are given by

$E_{vib}=(v+\frac{1}{2})h \sqrt{\frac{k}{\mu}} \\ =(v+ \frac{1}{2})(\frac{6.626 \times 10^{-34}J \;s}{2 \pi}) \sqrt{ \frac{1177 \;N/m}{8(1.66 \times 10^{-27} \;kg)} } \\ = (v+ \frac{1}{2})(3.14 \times 10^{-20}\;J)( \frac{1 \;eV}{1.602 \times 10^{-19} \;J} ) \\ = (v+ \frac{1}{2})(0.196 \;eV)$

For v=0,1,2

$E_{vib}=0.098 \;eV, 0.294 \;eV, 0.490 \;eV$