# Oscillations of Covalent Molecules

• BlueDevil14
In summary, the conversation is about the force of covalent bonds between diatomic molecules and finding the force constant for small oscillations around equilibrium using Hooke's Law and the Taylor series expansion of the force equation. The force constant is found to be A*b=579.15 N/m.
BlueDevil14

## 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.

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.

## 1. What are oscillations of covalent molecules?

Oscillations of covalent molecules refer to the vibrations or movements of atoms within a molecule. These oscillations are caused by the sharing of electrons between atoms in a covalent bond.

## 2. How do these oscillations affect the properties of covalent molecules?

The frequency and amplitude of these oscillations can affect the stability, reactivity, and other physical and chemical properties of covalent molecules. For example, molecules with higher oscillation frequencies tend to be more reactive.

## 3. What factors influence the oscillation frequency of covalent molecules?

The oscillation frequency of a covalent molecule is influenced by the strength of the covalent bond, the masses of the atoms involved, and the bond length. It can also be affected by external factors such as temperature and pressure.

## 4. How can we measure the oscillation frequency of covalent molecules?

The oscillation frequency of covalent molecules can be measured using techniques such as infrared spectroscopy, Raman spectroscopy, and nuclear magnetic resonance (NMR) spectroscopy. These methods involve exciting the molecules and measuring the energy absorbed or emitted during their oscillations.

## 5. What is the significance of studying oscillations of covalent molecules?

Understanding the oscillations of covalent molecules is important in fields such as chemistry, materials science, and biochemistry. It allows us to predict and control the properties and behavior of molecules, which is crucial in developing new materials and drugs, as well as understanding chemical reactions and biological processes.

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