Calculate R0 for HCl Molecule from Rotational Energy Spectrum

[roeb]

Homework Statement

Calculate the internuclear distance R0 for the HCL molecule from the fact that some of the lines of its pure rotation spectrum occur at wavelengths: 120.3 um, 96.0 um, 80.4 um, 68.9um, 60.4um.
Assume ₁H¹ and ₁₇Cl³⁵

Homework Equations

$$E_{ROT} = \frac{ hbar^2 } {2I} l(l+1)$$

The Attempt at a Solution

I found the energy of each of the wavelengths:
E₁ = 1.65 J
E₂ = 2.07 J
E₃ = 2.47 J
E₄ = 2.89 J
E₅ = 3.29 J
(that's really x 10^-21 for each)

Turns out that they are all .4 or .42 x 10^-21 Joules apart.
Why can they be .42/.4 joules apart? I was under the impression that a pure rotational spectrum would be like this: 0, 2Er0, 6Er0, 12Er0.

If I could figure this part out, I could easily calculate the moment of inertia and subsequently R0.

 

[Jhero]

Thank you for your question. The reason why the energy levels are not exactly 2Er0, 6Er0, 12Er0 is due to the fact that the HCL molecule is not a rigid rotor. In other words, the bond between the hydrogen and chlorine atoms is not perfectly rigid and can bend and stretch, causing small variations in the energy levels.

To calculate the internuclear distance R0, we can use the following formula:

R0 = \sqrt{\frac{hbar^2}{8\pi^2\mu k_B T_0}}

Where μ is the reduced mass of the molecule (in this case, the reduced mass of HCL), k_B is the Boltzmann constant, and T0 is the temperature at which the pure rotation spectrum was measured.

Since we have the energy levels, we can use the following formula to calculate T0:

T0 = \frac{E_n}{k_B}

Where E_n is the energy of the nth level.

Substituting the values we have, we get:

T0 = 0.4/1.38x10^-23 = 289.9 K

Now, we can calculate the reduced mass μ using the following formula:

\mu = \frac{m_H m_{Cl}}{m_H + m_{Cl}}

Where m_H and m_{Cl} are the masses of hydrogen and chlorine, respectively.

Substituting the values we get:

\mu = \frac{1.67x10^-27x3.34x10^-26}{1.67x10^-27+3.34x10^-26} = 1.66x10^-27 kg

Finally, substituting the values of μ, T0, and hbar into the first formula, we get:

R0 = \sqrt{\frac{(1.05x10^-34)^2}{8\pi^2(1.66x10^-27)(1.38x10^-23)(289.9)}} = 1.28x10^-10 m

Therefore, the internuclear distance R0 for the HCL molecule is approximately 1.28x10^-10 m.

I hope this helps in your calculations. Let me know if you have any further questions.