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

  • Thread starter bmarson123
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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 = [itex]\mu[/itex]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:

ehild

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

Thank you!

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

and to find [itex]\Delta[/itex]E, [itex]\Delta[/itex]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?
 

ehild

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Last edited:
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
 

ehild

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

Can you show your calculations in detail?

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

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

[itex]\Delta[/itex]E = hf = h x 1.12 x1010 = 7.4256 x10-24

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

r = [itex]\sqrt{1.520x10-45/12.314}[/itex]
= 1.1 x10-23m
 

ehild

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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?
 
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.
 

ehild

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

6.02 x 1023?

Do I need to divide the value of [itex]\mu[/itex] by that?
 

ehild

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

6.02 x 1023?

Do I need to divide the value of [itex]\mu[/itex] by that?
Yes.

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

Is the right answer 8.62 x 10-12m?
 

ehild

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

2.72 x10-10m?
 

ehild

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

It is OK.

ehild
 

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