Answer:
the energy of the third excited rotational state ![\mathbf{E_3 = 16.041 \ meV}](https://tex.z-dn.net/?f=%5Cmathbf%7BE_3%20%3D%2016.041%20%5C%20meV%7D)
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 = ![\dfrac{m_1 \times m_2}{m_1 + m_2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bm_1%20%5Ctimes%20m_2%7D%7Bm_1%20%2B%20m_2%7D)
μ = ![\dfrac{1 \times 35}{1 + 35}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%20%5Ctimes%2035%7D%7B1%20%2B%2035%7D)
μ = ![\dfrac{35}{36}](https://tex.z-dn.net/?f=%5Cdfrac%7B35%7D%7B36%7D)
∵ 1 μ = 1.66 × 10⁻²⁷ kg
μ = ![\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg](https://tex.z-dn.net/?f=%5C%5C%20%5C%5C%20%5Cdfrac%7B35%7D%7B36%7D%20%5Ctimes%201.66%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20%5C%20%20kg)
μ = 1.6139 × 10⁻²⁷ kg
![r_o = 127 \ pm = 127*10^{-12} \ m](https://tex.z-dn.net/?f=r_o%20%3D%20127%20%5C%20pm%20%3D%20127%2A10%5E%7B-12%7D%20%5C%20m)
The rotational level Energy can be expressed by the equation:
![E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)](https://tex.z-dn.net/?f=E_J%20%3D%20%5Cdfrac%7Bh%5E2%7D%7B8%20%5Cpi%5E2%20I%20%7D%20%5Ctimes%20J%20%28%20J%20%2B1%29)
where ;
J = 3 ( i.e third excited state) &
![I = \mu r^2_o](https://tex.z-dn.net/?f=I%20%3D%20%5Cmu%20r%5E2_o)
![E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)](https://tex.z-dn.net/?f=E_J%3D%20%5Cdfrac%7Bh%5E2%7D%7B8%20%20%5Cpi%20%20%5Cmu%20r%5E%202%20%5Cmur_o%20%7D%20%5Ctimes%20J%20%28%20J%20%2B1%29)
![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)](https://tex.z-dn.net/?f=E_3%20%3D%20%5Cdfrac%7B%286.63%20%5Ctimes%2010%5E%7B-34%7D%29%5E2%7D%7B8%20%20%5Ctimes%20%20%5Cpi%20%5E2%20%20%5Ctimes%201.6139%20%5Ctimes%2010%5E%7B-27%7D%20%5Ctimes%28%20127%20%5Ctimes%2010%5E%7B-12%7D%29%20%5E%202%20%20%7D%20%5Ctimes%203%20%28%203%20%2B1%29)
![E_3= 2.5665 \times 10^{-21} \ J](https://tex.z-dn.net/?f=E_3%3D%202.5665%20%5Ctimes%2010%5E%7B-21%7D%20%5C%20J)
We know that :
1 J = ![\dfrac{1}{1.6 \times 10^{-19}}eV](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7DeV)
![E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV](https://tex.z-dn.net/?f=E_3%3D%20%5Cdfrac%7B2.5665%20%5Ctimes%2010%5E%7B-21%7D%20%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7DeV)
![E_3 = 16.041 \times 10 ^{-3} \ eV](https://tex.z-dn.net/?f=E_3%20%3D%2016.041%20%20%5Ctimes%2010%20%5E%7B-3%7D%20%5C%20eV)
![\mathbf{E_3 = 16.041 \ meV}](https://tex.z-dn.net/?f=%5Cmathbf%7BE_3%20%3D%2016.041%20%5C%20meV%7D)
P=18000000/6 zeros. not sure how to do rest
Answer: A golfer hitting a golf ball.
Explanation:
The atomic particles move more in this option than the others.