The potential energy shared by two atoms in a diatomic molecule, depicted in Figure 17, is often approximated by the fairly simple function U(x)=(ax12)-(bx6)where constants a and b depend on the atoms involved. In Section 7, it is said that near its minimum value, it can be approximated by an even simpler function—it should “look like” a parabola. (a) In terms ofa and b, find the minimum potential energy U (x0) and the separation x0 at which it occurs. (b) The parabolic approximation UP(x)=U(xo)+12κ(x-xo)2has the same minimum value at x0 and the same first derivative there (i.e., 0). Its second derivative is k , the spring constant of this Hooke’s law potential energy. In terms of a and b, what is the spring constant of U (x)?

Short Answer

Expert verified

a. Minimum potential energy is Uxo=-b24aat x0=2ab16separation.

(b) The double derivative of U (x) at x0 gives spring constant of U(x) i.e.,κ=14b7a413

Step by step solution

01

Take first derivative of and equate it to zero

(a)

Taking derivative and equating to zero,

dUxdx=0 ddxax12-bx6=0 -12ax13+6bx7=0 xo=2ab16

This implies that at xo=2ab16the U(x) is minimum. Therefore, the minimum potential energy is

Uxo=axo12-bxo6=a2ab126-b2ab66=b2a4a2-b22a=-b24a

02

Take double derivative of U (x) at x0 and solve

(b)

Determine the second derivate of the potential for the separation x.

dUxdx=-12ax13+6bx7d2Uxdx2=156ax14-42bx8d2Uxdx2xo=156a2ab146-42b2ab86d2Uxdx2xo14b7a413

The parabolic approximation expression gives spring constant on taking double derivative at x0. Therefore the double derivative of U (x) at x0 gives spring constant of U (x) .

κ=14b7a413

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