Calculus-based Physics 1 & (PH 201-202) Assignment BACKNEXT FULL NEXT Chapter 08

Question

Calculus-based Physics 1 & (PH 201-202) Assignment BACKNEXT FULL NEXT Chapter 08, Problem 040 The potential energy of a diatomic molecule (a two-atom system like H2 or 02) is given by 126 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 (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation? (a) req

Explanation / Answer

If the potential energy function is:

U(r) = A/(r^12) - B/(r^6)

and the molecule is NOT ROTATING, the force between the two atoms is:

F(r) = -dU/dr = 12A/(r^13) - 6B/(r^7)

a) Equilibrium occurs when F = 0

0 = F(R) = 12A/(R^13) - 6B/(R^7)

= [-(6B*R^6) + (12A)]/(R^13)

= (6/(R^13))(2A - B*R^6)

=> R = (2A/B)^(1/6)

b) F(r) = (6/(r^13))(2A - B*r^6)

(dF/dR)(R) = (6*(-13)/R^14)(2A - B*R^6) + (6/(R^13))(-6B*R^5)

= 0 - 36B/R^8

= -36B/R^8

Therefore, since F(R) = 0, for r slightly less than R, F(r) is > 0: repulsive.

c) Likewise, for r slightly greater than R, F(r) is < 0: attractive.

Thus, the molecule "resists" being moved from its equilibrium point - as expected.

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