CALC 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 X 10-10 m and the dissociation energy is 1.54 X 10-18 J per molecule. Find the values of the constants a and b.

Solution 39E Problem (a) Step 1: Potential energy of two atoms is given by a b Ur r12 - r6 -----(1) r - interatomic space a and b - positive constants Step 2: To find F r Force is expressed as dU Fr - dr Fr - d ( 12 - 6 ) dr r r 12a 6b Fr r13 - r7 -----(2) Step 3: To plot U vs r r Step 4: To plot F vrr Problem (b) Step 1: To find the equilibrium distance between two atoms At equilibrium force is zero and potential energy is also zero. Therefore 12a 6b Fr r13 - r7 = 0 Step 2: Solving the above equation 2a 1/6 We get r = ( b ) This distance is equilibrium distance r 0 At distance r = ( 2a ) , the potential energy is minimum and hence the atoms are in 0 b stable equilibrium. Problem (c) Step 1: The atoms are at equilibrium distance, the minimum energy to disassociate the atoms to infinite distance is found using following conditions, U = 12 - b6 r r r r = r = ( 2a ) ----(3) 0 b Step 2: Solving the above two equations U = a - b (2b)2 (2b) ab b 2 U = 4a 2 - 2a 2 U = - b 4a Step 3: At infinite distance, the potential energy is zero. b2 dU = 4a -----(4) Hence the energy added (dU) must be equal to b 2 , so that the potential energy would 4a be zero. Problem (d) Step 1: Consider CO molecule -10 Distance between C and O atoms r = 1.13x10 o Dissociation energy dU = 1.54x10 J -18 To find a and b Step 2: We know that 2a 1/6 r0= ( b ) ----(3) Substituting r =o.13x10 m in (3) (2a ) = 1.13x10 m -10 b 2a -60 6 b = 2.08x10 m ----(5) Step 3: From (4) 2 dU = b = 1.54x10 J -----(6) 4a Taking reciprocal of equation (5) 2a = (2.08x10 -60)-1 b 59 -6 2a = 4.8x 10 m 4a = 2.4x 10 m ----(7)