Assignment 1Question 1:

The figures below show the variation of potential energy and interatomic force with the interatomic distance. The data is of two Argon atoms.

The figure below shows the potential energy plot as a function of distance, and the various plots are at different external forces, as the external forces are increased the graph shifts towards bottom.

increasing f ext∅ (r )

f (r )

r-

∅ (r ) in meV

The figure below shows the behavior of potential well as a function of equilibrium distance, initially at vary small distance from equilibrium point the it resembles the spring behavior and the springs constant (k) we can obtain from expanding the potential energy equation and then taking the coefficient of the linear term after taking the first derivative, which comes out to be

k=d2∅ /d r2=72Ebr02

Question 2:

Questions from the books:

f - in (N)

r (m)-

Q1: For the Lennard- Jones potential, in the presence of an external force, find the

expression for the new equilibrium point r0' as a function of f ext, and the maximum force

for which an equilibrium point still exists. Draw a plot showing the dis placement from

equilibrium δ=r 0'−r0 as a function of f ext, showing the linear region and where the

linear approximation breaks down.

Lennard- Jones potential: ∅ (r )=−Ar6

+ Br12

Where: A=2r 06Eb

B=r 012 Eb

r0= Equilibrium Distance

Eb= Binding Energy

Assuming linear behaviour around the equilibrium point,

δ=r 0'−r0=

1

(d2∅ ¿¿dr 2) f ext ¿ --------(1)

d2∅ /dr2=72Ebr02

So from we equation 1 we can get the new equilibrium distancer0'. But as the load

increases this linear approximation will not be valid, so the graph below is plotted comparing the distance obtained solving the Taylor’s equation with quadratic term too.

The graph in the green is solution of quadratic equation whereas the graph in blue is solution of linear equation.

The force after which the equilibrium point is not possible is f lim ¿¿= 7E-12 N.

And in the graph as we can see that the linear behaviour holds good for the loads upto 1E-12 N after that the error becomes more.

Q2: From Lennard- Jones potential:

φ (r )=−Ar6

+ Br12

WhereA=2r 06Eb, B=r 0

12 Eb and r0=(2 AB )16 = equilibrium distance.

Now force is defined as:

f (r )=−∂φ∂r

⇒ f (r )=−(6 Ar7−12 Br13 )=−6A

r7+12

B

r13

Now for finding the maximum value of force,

df (r )dr

=0

The above condition should be satisfied.

df (r )dr

=42 Ar8

−156 Br 14

=0

⇒r 6=267BA

⇒r=r1=[ 137 (2 BA )]16

Force ---

r0'

⇒r1=( 137 )16 r0=1.109 r 0

Substituting above expression of r1 in the expression of expression of force f(r), we get the required expression of maximum force.

f ( r1 )=−62 r0

6 Eb

(1.109 r0 )7+12

r012Eb

(1.109 r0 )13

∴ f (r1 )=−8.943Ebr0

Q3:

∅ (r )=−k q1q2r

+ Ar12

Where k = Coulomb’s Constant

At equilibrium:

k q1q2r02 −12× A

r013=0

A=k q1q2 r0

11

12

So, ∅ (r )=−k q1q2r

+k q1q2r0

11

12 r12

∅ (r )=−k q1q2r [1− 1

12 ( r 0r )11]

Now Binding Energy:

Eb=∅ (∞)−∅ (r 0 )=11k q1q212r 0