Respuesta :
Answer
given,
distance between the nuclei of an O₂ molecule = 1.20 x 10⁻¹⁰ m
mass of oxygen atom = 2.66 x 10⁻²⁶ Kg
the reduced mass of O₂ molecule =
[tex]\mu=\dfrac{m_1m_2}{m_1+m_2}[/tex]
[tex]\mu=\dfrac{m_0 m_0}{m_0+m_0}[/tex]
[tex]\mu=\dfrac{m_0}{2}[/tex]
[tex]\mu=\dfrac{2.66 \times 10^{-26}}{2}[/tex]
[tex]\mu=1.33 \times 10^{-26}[/tex]
moment of inertia of O₂ molecule
[tex]I = \mu r^2[/tex]
[tex]I = 1.33 \times 10^{-26} \times (1.2\times 10^{-10})^2[/tex]
I = 1.9152 x 10⁻⁴⁶ kg.m²
a) Rotational energy of oxygen molecule
[tex]E_j = \dfrac{h^2}{2l}j(j+1)[/tex]
J = 0
[tex]E_j =0[/tex]
J = 1
[tex]E_1= \dfrac{h^2}{2l}(1)(1+1)[/tex]
[tex]E_1= \dfrac{h^2}{l}[/tex]
[tex]E_1= \dfrac{(1.055 \times 10^{-34})^2}{1.9152\times 10^{-46}}[/tex]
[tex]E_1=5.81\times 10^{-23}J[/tex]
[tex]E_1=\dfrac{5.81\times 10^{-23}J}{1.6\times 10^{-19}}[/tex]
E₁ = 3.63 x 10⁻⁴ eV
J = 2
[tex]E_2= \dfrac{h^2}{2l}(2)(2+1)[/tex]
[tex]E_2= 3\dfrac{h^2}{l}[/tex]
[tex]E_2= 3\times 3.36 \times 10^{-4}[/tex]
E₂ = 1.089 x 10⁻³ eV
b) Effective force constant between the molecule
[tex]E = (v+\dfrac{1}{2})\dfrac{h}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
for v = 0
[tex]E =\dfrac{h}{4\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]E =\dfrac{1.055\times 10^{-34}}{4\pi}\sqrt{\dfrac{1177}{2.66\times 10^{-26}}}[/tex]
E = 1.569 x 10⁻²¹ J
[tex]E = \dfrac{1.569 \times 10^{-21}}{1.6\times 10^{-19}}[/tex]
E₀ = 9.8 x 10⁻³ eV
for v = 1
E₁ = 3 E₀
E₁ = 3 x 9.8 x 10⁻³
E₁ = 29.4 x 10⁻³ eV
For v = 2
E₂ = 5 E₀
E₂ = 5 x 9.8 x 10⁻³
E₂ = 49 x 10⁻³ eV
