Question

The potential energy of two atoms in a diatomic molecule is approximated by U(r)=(a/r12)−(b/r6), where r is the spacing between atoms and a and b are positive constants. (a) Find the force F(r) on one atom as a function of r. Draw two graphs: one of U(r) versus r and one of F(r) versus r. (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) 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? This is called the dissociation energy of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13×10−10m and the dissociation energy is 1.54×10−18J per molecule. Find the values of the constants a and b.

Answer 1

Answer:

A) F(r) = [12a/(r^(13))] - [6b/(r^(7))]

ii) Graphs are attached

B) Equilibrium Distance = (2a/b)^(1/6)

C) Minimum Energy = b²/4a

D) a = 6.67 x 10^(-138) Jm^(12)

b = 6.41 x 10^(-78) Jm^(6)

Explanation:

I've attached the explanation of A-C alongside the graphs

D) i) From the question, we are to make r equal to the derivative of "r" we got when F(r) = 0 which was r = (2a/b)^(1/6)

Thus, since we are given equilibrium distance as: 1.13 x 10^(-10), hence;

(2a/b)^(1/6) = 1.13 x 10^(-10)m

So, (2a/b)= [1.13 x 10^(-10)]^(6)

a = (b/2)[1.13 x 10^(-10)]^(6)

From earlier, we saw that b²/4a = U(r)

Thus since U(r) = 1.54 x 10^(-18) from the question, b²/4a = 1.54 x 10^(-18)

Putting a = (b/2)[1.13 x 10^(-10)]^(6);

We have;

(b²) / [(4b/2)[1.13 x 10^(-10)]^(6)]] = 1.54 x 10^(-18)

b/2 = [1.13 x 10^(-10)]^(6)] x 1.54 x 10^(-18)

So, b = 6.41 x 10^(-78) Jm^(6)

ii) Putting (6.41 x 10^(-78))² for b in;

a = (b/2)[1.13 x 10^(-10)]^(6)

We have, a = (6.41 x 10^(-78))²/ ( 4 x 1.54 x 10^(-18)

So a = 6.67 x 10^(-138) Jm^(12)