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Equilibrium separation 
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#1
Oct305, 08:22 PM

P: 17

Oh boy...this problem has stumped me. I am not even sure I know where to start:
The potential energy of a diatomic molecule (a twoatom system like H2 or O2) is given by U= (A/r^12)  (B/r^6) where r is the separation of the two atoms of the molecule and A and B are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation  that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (atoms are pushed apart) or attractive (atoms are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation? Okay. I started out thinking that because we are looking for the force, we would need to use F(x) =  (d/dx)(U(x)), and after differentiation F(x) = (12A/r^13) + (6B/r^7) So here is the big "now what?" Please help. Thank you  any help is greatly appreciated. ~Gin 


#2
Oct305, 08:34 PM

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#3
Oct305, 09:27 PM

P: 17

okay...but there are three unknowns in my equation.... Is there a step I am leaving out?? Even solving algebraically only reduces down to 2ABr^6 = 0. Does something cancel out that I am not aware of? Or maybe we are supposed to be left with an equation?



#4
Oct305, 09:30 PM

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Equilibrium separation
You are told "A and B are positive constants". So, you can expect that your answer is in terms of those positive [but otherwise unspecified] constants.



#5
Oct305, 10:01 PM

P: 17

Ahhh...so I got my equation down to (2A/B)^(1/6) = r. So that value of r is the equilibrium separation. Now for the next part, figuring out atomic behaviors based on larger/smaller values in relation to equilibrium sep. This should be interesting.
Equation: 2ABr^6=0 If sep. is larger than equilibrium, we should have a negative force, and if it is smaller than eq., we should have a positive force. I think a negative force will draw the atoms together (?) Any thoughts are appreciated. Thank ya'll. gin 


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