# Homework Help: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium sep

1. Aug 11, 2011

### bmarson123

Problem:
The adjacent lines in the pure rotational spectrum of 35CI19F are separated by a frequency of
1.12 x 1010 Hz. What is the equilibrium separation of the atoms in this molecule?

The attempt at a solution

So, the equation I think I need to use to find the equilibrium separation is I = $\mu$r02
But I don't know how I can find the inertia,as I don't know how the frequency and inertia are linked?

Last edited: Aug 11, 2011
2. Aug 11, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

The emitted or absorbed photon energies differ by ΔE=[STRIKE]h[/STRIKE]2/I. How is the photon energy related to frequency?

I is the moment of inertia of the diatomic molecule: I=r2m1m2/(m1+m2), where r is the distance between the atoms.

ehild

3. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Thank you!

Ok, so I've tried using $\Delta$E = [STRIKE]h[/STRIKE]/I

and to find $\Delta$E, $\Delta$E = hf

and then just number crunching that all through, using m1 = 35 and m2 = 19,

I get r = 1.1 x 10-6m

Is that right? It seems pretty big?

4. Aug 12, 2011

### ehild

Last edited: Aug 12, 2011
5. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Yes, does help if I look at my notes properly.

Working that through I get r = 1.1 x 10-23m

6. Aug 12, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Can you show your calculations in detail?

ehild

7. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

[STRIKE]h[/STRIKE] = h / 2$\pi$ = 1.055 x10-34 Js

$\Delta$E = hf = h x 1.12 x1010 = 7.4256 x10-24

I = [STRIKE]h[/STRIKE]2/$\Delta$E
= 1.113 x10-68 / 7.4256 x10-24
= 1.520 x 10-45

r = $\sqrt{1.520x10-45/12.314}$
= 1.1 x10-23m

8. Aug 12, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

What is the reduced mass? Did you take it 12.34 kg? Is the mass of a molecule in the range of kg-s?

9. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

That's what I've taken it as, yes. I realise now how stupid that is but I don't know how to work out the mass from the numbers. I don't even know what the numbers are next to the elements.

10. Aug 12, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Those numbers are the "molar" masses, the mass of one mole of atoms in gram unit. How many atoms are there in one mole?

ehild

11. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

6.02 x 1023?

Do I need to divide the value of $\mu$ by that?

12. Aug 12, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Yes.

ehild

13. Aug 12, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Is the right answer 8.62 x 10-12m?

14. Aug 12, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

Not yet. 12.34 is the molar mass in grams, so divided by the Avogadro number, you get μ in grams, but you need it in kg-s in the equation.

ehild

15. Aug 13, 2011

### bmarson123

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

2.72 x10-10m?

16. Aug 13, 2011

### ehild

Re: Molecular Rotation - Adjacent lines in pure rotational spectrum: find equilibrium

It is OK.

ehild