The potential energy of two atoms in a diatomic molecule is approximated by U (r) = a/r^1/2-b/r^6, 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.

b. Find the equilibrium distance between the two atoms. Express your answer in terms of the variables a and b. 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, U (r=[infinity]) = 0? This is called the dissociation energy of the molecule. Express your answer in terms of the variables a and b.

d. For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 x 10^-10m and the dissociation energy is 1.54 X 10^-18J per molecule. Find the value of the constant a. Find the value of the constant b.

Question

Physics

Eli Shaw

2 September, 14:39

The potential energy of two atoms in a diatomic molecule is approximated by U (r) = a/r^1/2-b/r^6, 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.

b. Find the equilibrium distance between the two atoms. Express your answer in terms of the variables a and b. 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, U (r=[infinity]) = 0? This is called the dissociation energy of the molecule. Express your answer in terms of the variables a and b.

d. For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 x 10^-10m and the dissociation energy is 1.54 X 10^-18J per molecule. Find the value of the constant a. Find the value of the constant b.

+3

Answers (1)

Know the Answer?

Tiara Warren · Answer 1

a. F (r) = 12ar^-13 - 6br-7

b. U (r) = - b²/4a.

c. b²/4a

d. a = 6.67E-138J, b = 6.41E-78J

Explanation:

Given

U (r) = a/r^12 - b/r^6

a. Calculating Force, F (r)

F (r) = - dU/dr

dU/dr = ar^-12 - br^-6

dU/dr = - 12ar^-13 + 6br-7

F (r) = - dU/dr

F (r) = 12ar^-13 - 6br-7

b. Finding the equilibrium distance;

Summation F = 0

First, we solve for r

F (r) = 12ar^-13 - 6br-7 = 0

12ar^-13 = 6br-7 - - - Multiply both sides by r^7

12ar^-6 = 6b

r^-6 = 12a/6b

r^-6 = 2a/b

r = (2a/b) ^ - (1/6) ...

U (r) = a/r^12 - b/r^6

U (r) = a / ((b/2a) ^ - (1/6)) ^12 - a / ((b/2a) ^ - (1/6)) ^6

U (r) = a / (2a/b) ² - b / (2a/b)

U (r) = a/4a²/b² - b/2a/b

U (r = ab²/4a² - b²/2a

U (r) = b²/4a - b²/2a

U (r) = (b² - 2b²) / 4a

U (r) = - b²/4a.

c.

At Equilibrium distance, (r) = 1.13 * 10^-10

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

(2a/b) = (1.13 * 10^-10) ^6

2a = ((1.13 * 10^-10) ^6) b

a = ½ ((1.13 * 10^-10) ^6) b

d. Dissociation Energy

b²/4a = 1.54 * 10^-18

b²/4 (½ ((1.13 * 10^-10) ^6) b) = 1.54 * 10^-18

b = 1.54 * 10^-18 * 2 ((1.13 * 10^-10) ^6))

b = 6.41E-78J

Solving for a

a = ½ ((1.13 * 10^-10) ^6) * 6.41E-78J

a = 6.67E-138J