Question

The potential energy of two atoms in a diatomic molecule is approximated by U(r)=d/r12-b/r6, where r is the spacing between atoms and a and b are positive constants. Find the force F(r) on one atom as a function of r. Find the equilibrium distance between the two atoms. Express your answer in terms of the variables a and b. Is this equilibrium stable? Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart, U(r=infinity) = 0? This is called the dissociation energy of the molecule. Express your answer in terms of the variables a and b. For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 times 10-10m and the dissociation energy is 1.54 times 10-18J per molecule. Find the value of the constant a. Find the value of the constant b.

Answer 1

Concepts and reason

This problem is based on the concept of conversion of potential energy to force on atom and equilibrium distance between two atoms.

Initially, differentiate the potential energy expression to calculate the expression for the force with a negative sign, calculate the equilibrium distance between atoms.

Later, if potential energy is written in an expression, the differential of the potential energy with negative sign gives the force on the atom.

Finally, equilibrium distance is measured by using the expression of the force on the atom, this expression is equaled to zero to find the equilibrium distance.

Fundamentals

The force $F(r)$ is written as follows:

$F(r) = - \frac{d}{{dr}}\left[ {U(r)} \right]$

Here, $U(r)$ is the function of potential energy.

The force $F(r)$ should be zero, therefore it is written as follows:

$\begin{array}{c}\\F(r) = \left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right]\\\\ = 0\\\end{array}$

Here, $r$ is the spacing between atoms, $a$ and $b$ are positive constants.

The case for stability is given as,

$\frac{{dF(r)}}{{dr}} < 0$

Here, $F(r)$ is the function of force.

(a)

The expression for the potential energy of two atoms in a diatomic molecule is approximated as,

$U(r) = \frac{a}{{{r^{12}}}} - \frac{b}{{{r^6}}}$ …… (1)

The force $F(r)$ is written as,

$F(r) = - \frac{d}{{dr}}\left[ {U(r)} \right]$ …… (2)

From expression (1) and (2),

$\begin{array}{c}\\F(r) = - \frac{d}{{dr}}\left[ {\frac{a}{{{r^{12}}}} - \frac{b}{{{r^6}}}} \right]\\\\ = - \left[ {\frac{{ - 12a}}{{{r^{13}}}} - \frac{{ - 6b}}{{{r^7}}}} \right]\\\\ = \left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right]\\\end{array}$

(b)

The equilibrium distance between two atoms is calculated as follows:

$\begin{array}{c}\\F(r) = \left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right]\\\\ = 0\\\end{array}$

Further, it is written as follows:

$\begin{array}{l}\\\left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right] = 0\\\\\frac{{12a}}{{{r^{13}}}} = \frac{{6b}}{{{r^7}}}\\\\{r^6} = \frac{{2a}}{b}\\\end{array}$

Solve for r.

$r = {\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}$

(c)

The case for stability is given as,

$\frac{{dF(r)}}{{dr}} < 0$

The differentiation of $F(r)$ with respect to $r$ is written as follows:

$\begin{array}{c}\\\frac{{dF(r)}}{{dr}} = \frac{d}{{dr}}\left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right]\\\\ = \left[ {\frac{{ - \left( {12} \right)\left( {13} \right)a}}{{{r^{14}}}} + \frac{{\left( 6 \right)\left( 7 \right)b}}{{{r^8}}}} \right]\\\\ = \left[ {\frac{{ - 156a}}{{{r^{14}}}} + \frac{{42b}}{{{r^8}}}} \right]\\\end{array}$

Substitute the value ${\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}$ for $r$ in above expression.

$\begin{array}{c}\\\frac{{dF(r)}}{{dr}} = \left[ {\frac{{ - 156a}}{{{{\left( {{{\left( {\frac{{2a}}{b}} \right)}^{\frac{1}{6}}}} \right)}^{14}}}} + \frac{{42b}}{{{{\left( {{{\left( {\frac{{2a}}{b}} \right)}^{\frac{1}{6}}}} \right)}^8}}}} \right]\\\\ = \left[ {\frac{{ - 156a}}{{{{\left( {\frac{{2a}}{b}} \right)}^{\frac{{14}}{6}}}}} + \frac{{42b}}{{{{\left( {\frac{{2a}}{b}} \right)}^{\frac{8}{6}}}}}} \right]\\\\ < 0\\\end{array}$

(d)

Substitute $- a{\left( {\frac{{2a}}{b}} \right)^{ - 2}}$ for ${U_{{\rm{final}}}}$ and $b{\left( {\frac{{2a}}{b}} \right)^{ - 1}}$ for ${U_{{\rm{initial}}}}$ in the equation ${U_{{\rm{added}}}} = {U_{{\rm{final}}}} - {U_{{\rm{initial}}}}$ .

$\begin{array}{c}\\{U_{{\rm{added}}}} = - a{\left( {\frac{{2a}}{b}} \right)^{ - 2}} - b{\left( {\frac{{2a}}{b}} \right)^{ - 1}}\\\\ = - \frac{{{b^2}}}{{4a}} + \frac{{{b^2}}}{{2a}}\\\\ = \frac{{{b^2}}}{{4a}}\\\end{array}$

(e)

The value of the aquarium distance for the Co molecule is equal to,

$\begin{array}{c}\\{r_0} = {\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}\\\\ = 1.13 \times {10^{ - 10}}{\rm{ m}}\\\end{array}$

And the dissociation energy is as follows:

$\begin{array}{c}\\{U_{{\rm{added}}}} = \frac{{{b^2}}}{{4a}}\\\\ = 1.54 \times {10^{ - 18}}{\rm{ }}\\\end{array}$

Therefore, from the equation ${r_0} = {\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}$ and $\frac{{{b^2}}}{{4a}}$ ,

$\frac{{2a}}{b} = 2.08 \times {10^{ - 60}}$

Using the equation $\frac{{{b^2}}}{{4a}} = 1.54 \times {10^{ - 18}}{\rm{ m}}$ .

${b^2} = 4a\left( {1.54 \times {{10}^{ - 18}}{\rm{ m}}} \right)$

Solve for a.

$a = 6.656 \times {10^{ - 138}}$

(f)

Using the equation $\frac{{{b^2}}}{{4a}} = 1.54 \times {10^{ - 18}}{\rm{ m}}$ .

${b^2} = 4a\left( {1.54 \times {{10}^{ - 18}}{\rm{ m}}} \right)$

Substitute $6.656 \times {10^{ - 138}}$ for a.

$\begin{array}{c}\\{b^2} = 4\left( {6.656 \times {{10}^{ - 138}}} \right)\left( {1.54 \times {{10}^{ - 18}}{\rm{ m}}} \right)\\\\b = \sqrt {4\left( {6.656 \times {{10}^{ - 138}}} \right)\left( {1.54 \times {{10}^{ - 18}}{\rm{ m}}} \right)} \\\\ = 6.40 \times {10^{ - 78}}\\\end{array}$

Ans: Part a

The force $F(r)$ on one atom as a function of $r$ is $\left[ {\frac{{12a}}{{{r^{13}}}} - \frac{{6b}}{{{r^7}}}} \right]$ .

Part b

The equilibrium distance between the two atoms is ${\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}}$ .

Part c

This Equilibrium is stable.

Part d

The dissociation energy is equal to $\frac{{{b^2}}}{{4a}}$ .

Part e

The value of the constant a is equal to $6.656 \times {10^{ - 138}}$ .

Part f

The value of the constant b is equal to $6.40 \times {10^{ - 78}}$ .