Question

The potential energy of a system of two atoms is given by the expression:

𝑈 = − 𝐴 𝑟

2 ⁄ + 𝐵 𝑟

10 ⁄

A stable molecule is formed with release of 8.0 eV of energy when the interatomic distance is 

2.8 Å. 

i. Calculate A and B. [4 marks]

ii. Determine the force needed to dissociate this molecule into atoms and the interatomic 

distance at which the dissociation occurs. 

Answer 1

Answer

Energy is given

$E=8eV=12.8\times10^{-19}J$

Interatomic distance

$r=2.8A^°=2.8\times 10^{-10}m$

Potential energy is given

$U=-Ar^2+Br^{10}$

Now force ia given

$F=-\frac{dU}{dr}$

For equilibrium state

$F=0=-(-2Ar+10Br^9)$

By solving this

$r^8=\frac{A}{5B}$

$A=5Br^8$

$A=5\times (2.8\times10^{-10})^8B\\A=1.9\times10^{-76}B$

This relation shows value of A and B.

Dissociation energy is

$E=\frac{B^2}{4A}=\frac{B^2}{4\times1.9\times10^{-76}B}$

=$8\times1.6\times10^{-19}$ (Given)

After solving

B=51.2$\times10^{-95}$ J/m^2

Now putting the value of B in above equation

$A=1.9\times10^{-76}\times 51.2\times10^{-95}$

$A=93.28\times10^{-171}J/m^{10}$