Question

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

Answer 1

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

Step-by-step 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 =
`(m_1 * m_2)/(m_1 + m_2)`

μ =
`(1 * 35)/(1 + 35)`

μ =
`(35)/(36)`

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ =
`\\ \\ (35)/(36) * 1.66 * 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 = (h^2)/(8 \pi^2 I ) * J ( J +1)`

where ;

J = 3 ( i.e third excited state) &

`I = \mu r^2_o`

`E_J= (h^2)/(8 \pi \mu r^ 2 \mur_o ) * J ( J +1)`

`E_3 = ((6.63 * 10^(-34))^2)/(8 * \pi ^2 * 1.6139 * 10^(-27) *( 127 * 10^(-12)) ^ 2 ) * 3 ( 3 +1)`

`E_3= 2.5665 * 10^(-21) \ J`

We know that :

1 J =
`(1)/(1.6 * 10^(-19))eV`

`E_3= (2.5665 * 10^(-21) )/(1.6 * 10^(-19))eV`

`E_3 = 16.041 * 10 ^(-3) \ eV`

`\mathbf{E_3 = 16.041 \ meV}`