Answer:
the energy of the third excited rotational state [tex]\mathbf{E_3 = 16.041 \ meV}[/tex]
Explanation:
Given that :
hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm
Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.
Thus; the reduced mass μ = [tex]\dfrac{m_1 \times m_2}{m_1 + m_2}[/tex]
μ = [tex]\dfrac{1 \times 35}{1 + 35}[/tex]
μ = [tex]\dfrac{35}{36}[/tex]
∵ 1 μ = 1.66 × 10⁻²⁷ kg
μ = [tex]\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg[/tex]
μ = 1.6139 × 10⁻²⁷ kg
[tex]r_o = 127 \ pm = 127*10^{-12} \ m[/tex]
The rotational level Energy can be expressed by the equation:
[tex]E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)[/tex]
where ;
J = 3 ( i.e third excited state) &
[tex]I = \mu r^2_o[/tex]
[tex]E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)[/tex]
[tex]E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)[/tex]
[tex]E_3= 2.5665 \times 10^{-21} \ J[/tex]
We know that :
1 J = [tex]\dfrac{1}{1.6 \times 10^{-19}}eV[/tex]
[tex]E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV[/tex]
[tex]E_3 = 16.041 \times 10 ^{-3} \ eV[/tex]
[tex]\mathbf{E_3 = 16.041 \ meV}[/tex]
