Unravelling the Separation of 35CI19F Atoms: A Problem Solved

[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 = \mur₀²
But I don't know how I can find the inertia,as I don't know how the frequency and inertia are linked?

 

[ehild]

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

 

[bmarson123]

Thank you!

Ok, so I've tried using \DeltaE = [STRIKE]h[/STRIKE]/I

and to find \DeltaE, \DeltaE = 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]

Ok, so I've tried using \DeltaE = [STRIKE]h[/STRIKE]/I

It is \DeltaE = [STRIKE]h[/STRIKE]²/I. Read: http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotrig.html#c3

ehild

 

[bmarson123]

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

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

 

[ehild]

Can you show your calculations in detail?

ehild

 

[bmarson123]

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

\DeltaE = hf = h x 1.12 x10¹⁰ = 7.4256 x10^-24

I = [STRIKE]h[/STRIKE]²/\DeltaE
= 1.113 x10^-68 / 7.4256 x10^-24
= 1.520 x 10^-45

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

 

[ehild]

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

 

[bmarson123]

That's what I've taken it as, yes. I realize 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]

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

 

[bmarson123]

6.02 x 10²³?

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

 

[ehild]

6.02 x 10²³?

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

Yes.

ehild

 

[bmarson123]

Is the right answer 8.62 x 10^-12m?

 

[ehild]

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

 

[bmarson123]

2.72 x10^-10m?

 

[ehild]

It is OK.

ehild