Oscillations of Covalent Molecules

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SUMMARY

The discussion focuses on calculating the force constant for small oscillations of diatomic molecules bound by covalent bonds, specifically using the hydrogen molecule as an example. The force is described by the equation F_{r} = A[ e^{- 2b( r - R_0 )} - e^{ - b(r - R_0 )}], with constants A = 2.97 * 10^{-8} N, b = 1.95 \times 10^{10} m^{-1}, and R_0 = 7.4 \times 10^{-11} m. By applying the Taylor series expansion and retaining only the constant and linear terms, the force constant k is determined to be 579.15 N/m, derived from the relationship k = A*b.

PREREQUISITES
  • Understanding of covalent bonding and molecular interactions
  • Familiarity with Hooke's Law: F = -k*x
  • Knowledge of Taylor series expansion in calculus
  • Basic principles of molecular physics and oscillations
NEXT STEPS
  • Study the application of Taylor series in physics problems
  • Explore the relationship between force constants and molecular stability
  • Learn about the quantum mechanical treatment of diatomic molecules
  • Investigate the implications of covalent bond strength on molecular vibrations
USEFUL FOR

Students and educators in physics, particularly those focusing on molecular dynamics, quantum mechanics, and physical chemistry, will benefit from this discussion.

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
F_{r} = A[ e^{- 2b( r - R_0 )} - e^{ - b(r - R_0 )}]
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, 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}.

Find the force constant for small oscillations around equilibrium

Hint: Use the Taylor series expansion for e^x

i.e. e^{x}=1+x+\frac{x^{2}}{2}...

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?
 
Last edited:
for anyone else reading this thread, here is the Taylor expansion for the bracketed term (to the sixth power):

-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}...

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

k=A*b=579.15 N/m

I hope someone reads this and avoids all of my frustration
 
Last edited:
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.
 

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