Oscillations of Covalent Molecules

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BlueDevil14
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Homework Statement



Many diatomic (two-atom) molecules are bound together by covalent bonds that are much stronger than the van der Waals interaction. Experiment shows that for many such molecules, the interaction can be described by a force of the form
[itex]F_{r} = A[ e^{- 2b( r - R_0 )} - e^{ - b(r - R_0 )}][/itex]
where A and b are positive constants, r is the center-to-center separation of the atoms, and R_0 is the equilibrium separation. For the hydrogen molecule, [itex]A = 2.97 * 10^{ - 8} {\rm N}, b = 1.95 \times 10^{10} {\rm m}^{ - 1}, \text{and } R_0 = 7.4 \times 10^{ - 11} {\rm m}.[/itex]

Find the force constant for small oscillations around equilibrium

Hint: Use the Taylor series expansion for e^x

i.e. [itex]e^{x}=1+x+\frac{x^{2}}{2}...[/itex]

Homework Equations



Hooke's Law: F=-k*x


The Attempt at a Solution



I assume that the displacement for Hooke's Law is r/2 from the equation. We know force as a function of r already, and everything else is constant. The question is more math related, because I do not remember how to simplify this at all.
 
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I could really use some help. I am getting nowhere.
 
r-R0=Δr, Δr is the change of distance between the atoms. The bond is like a string, and the force between the atoms is of form F=-kΔr for small Δr-s, where k is the force constant. Use the Taylor-series expansion of exp(-2bΔr) and exp(-bΔr) in terms of Δr, and keep only the constant and linear terms, find k.

ehild
 
Thanks. Can you explain why I only keep the constant and linear terms?
 
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for anyone else reading this thread, here is the Taylor expansion for the bracketed term (to the sixth power):

[itex]-b Δr+\frac{3 b^2 Δr^2}{2}-\frac{7 b^3 Δr^3}{6}+\frac{5 b^4 Δr^4}{8}-\frac{31 b^5 Δr^5}{120}+\frac{7 b^6 Δr^6}{80}...[/itex]

Therefore Hooke's Law may be written as [itex]F_{r}=-AbΔr[/itex]

k=A*b=579.15 N/m

I hope someone reads this and avoids all of my frustration
 
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BlueDevil14 said:
Thanks. Can you explain why I only keep the constant and linear terms?
because the higher order terms are so small.

ehild
 
Thanks. It all makes sense now.