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andre [41]
3 years ago
9

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

up the molecule are hydrogen-1 (protium) and chlorine-35. (a) Find the energy of the third excited rotational state; that is, the J
Chemistry
1 answer:
natta225 [31]3 years ago
8 0

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

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}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

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)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

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