The potential energy of a diatomic molecule (for example, H 2 or O 2 ) is given

Question

The potential energy of a diatomic molecule (for example, H2 or O2) is given by: U = A/r12 - B/r6, where r is the separation distance between the atoms that make up the molecule and A and B are positive constants. This potential energy is due to the force that binds the atom together. If we assume A = 1.15 and B = 3.00, find the equilibrium separation, that is the distance between the atoms at which the force on each other is zero.

Is the force repulsive or attractive if their separation is smaller?

Explanation / Answer

Given that

The potential energy of a diatomic molecule (for example, H2 or O2) is given by

U = A/r12 - B/r6

And we know that

F(r) = -dU(r)/dr

In this case,

dU(r)/dr = [-(12A)/r13 ]+[ (6B)/r7 }]

F(r) = -dU(r)/dr = [(12A)/r13 ] - [(6B)/r7 ]

And now at the equilibrium, the potential is at minimum, that means the force is zero

Therefore F(r) =0

Now

0 = [(12A)/r13 ] - [(6B)/r7 ]

r6 =2A/B

Then r =(2A/B)1/6

          =(2*1.15/3)1/6

         =0.956

Now that is the distance between the atoms at which the force on each other is zero.Then the separation is becomes smaller then the force will be attractive.

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