Understanding Molecular Vibration of HF: Equilibrium Distance and Binding Energy

[schniefen]

Homework Statement

You estimate the potential energy for the interaction of a negatively charged fluorine ion and a proton as a function of distance $d$ by $V(d)=\frac{A}{d^4}-\frac{B}{d}$, where $A=2.4\cdot 10^{-28}$ Jm$^4$ and $B=0.6\cdot 10^{-58}$ Jm.
a) Calculate the equilibrium distance $a$ between the ions.
b) Calculate the binding energy between the ions.
c) Calculate the vibration frequency for the HF molecule (in one dimension) for small elongations from the equilibrium distance based on this model. Assume that the heavy fluorine ion is at rest and use the proton mass $m_p=1.6\cdot 10^{-27}$ kg.

Relevant Equations

I am a little unsure what the relevant equations are, hence the question. However, this exercise follows from lectures on oscillations and coupled oscillations, where molecules are modelled as spheres attached to springs. Maybe the potential and kinetic energy of a spring attached to a mass are relevant equations.

Starting with a), I have learnt that the potential energy function has a minimum at the equilibrium distance $a$. So at the equilibrium distance the derivative of the potential energy function should equal zero:

$\begin{align} V'(a)&=-\frac{4A}{a^5}+\frac{B}{a^2}=0 \nonumber \\ \iff Ba^5 &=4Aa^2 \nonumber \\ \iff a&=\left(\frac{4A}{B}\right)^{1/3} \nonumber \\ \end{align}$​

For b), I have learnt that the second derivative of the potential energy, evaluated at the equilibrium distance $a$, can be thought of as the spring constant $k$. So,

$k=V''(a)=\frac{20A}{a^6}-\frac{2B}{a^3}$
​

Though from hereon I am unsure how to continue.

​

 

[FinBurger]

So then the oscillation frequency omega will be sqrt(k/m). m is the mass of the proton (because the told you to consider the Fluorine ion to be so heavy it is assumed to be stationary. If they hadn't, then you should use the reduced mass).

As for the binding energy, evaluate the V function at the equilibrium point.

 

[DrClaude]

Homework Statement:: You estimate the potential energy for the interaction of a negatively charged fluorine ion and a proton as a function of distance $d$ by $V(d)=\frac{A}{d^4}-\frac{B}{d}$, where $A=2.4\cdot 10^{-28}$ Jm and $B=0.6\cdot 10^{-58}$ Jm$^4$.

The numerical value of the constants appear to be inverted.

Starting with a), I have learnt that the potential energy function has a minimum at the equilibrium distance $a$. So at the equilibrium distance the derivative of the potential energy function should equal zero:

$\begin{align} V'(a)&=-\frac{4A}{a^5}+\frac{B}{a^2}=0 \nonumber \\ \iff Ba^5 &=4Aa^2 \nonumber \\ \iff a&=\left(\frac{4A}{B}\right)^{1/3} \nonumber \\ \end{align}$​

Correct. You should probably also calculate the numerical value.

For b), I have learnt that the second derivative of the potential energy, evaluated at the equilibrium distance $a$, can be thought of as the spring constant $k$. So,

$k=V''(a)=\frac{20A}{a^6}-\frac{2B}{a^3}$
​

Though from hereon I am unsure how to continue.

​

This is actually the answer to c), you haven't done b).

 

[schniefen]

The numerical value of the constants appear to be inverted..

You were right, fixed it.

This is actually the answer to c), you haven't done b).

As @FinBurger pointed out, I just evaluate $V$ at the equilibrium distance for b). However, why at the equilibrium distance? Is this where the kinetic energy is $0$ and thus we obtain the total energy of the system?

 

[Steve4Physics]

You were right, fixed it.

As @FinBurger pointed out, I just evaluate $V$ at the equilibrium distance for b). However, why at the equilibrium distance? Is this where the kinetic energy is $0$ and thus we obtain the total energy of the system?

The binding energy is the minimum energy (increase in potential energy) needed to change from the equilibrium position to being completely separated. Hence:
Binding energy = V(∞) - V(a) = 0 - V(a) = -V(a)