Chapter 5: Q54E (page 190)

The potential energy shared by two atoms in a diatomic molecule, depicted in Figure 17, is often approximated by the fairly simple function $U (x) = (\frac{a}{x^{12}}) - (\frac{b}{x^{6}})$ 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 $U_{P} (x) = U (x_{o}) + \frac{1}{2} κ {(x - x_{o})}^{2}$ has 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 $U (x_{o}) = - \frac{b^{2}}{4 a}$ at $x_{0} = {(\frac{2 a}{b})}^{\frac{1}{6}}$ separation.

(b) The double derivative of U (x) at x₀ gives spring constant of U(x) i.e., $κ = 14 {(\frac{b^{7}}{a^{4}})}^{\frac{1}{3}}$

Step by step solution

Take first derivative of and equate it to zero

(a)

Taking derivative and equating to zero,

$\frac{d U (x)}{d x} = 0 \frac{d}{d x} (\frac{a}{x^{12}} - \frac{b}{x^{6}}) = 0 \frac{- 12 a}{x^{13}} + \frac{6 b}{x^{7}} = 0 x_{o} = {(\frac{2 a}{b})}^{\frac{1}{6}}$

This implies that at

x_{o} = {(\frac{2 a}{b})}^{\frac{1}{6}}

the U(x) is minimum. Therefore, the minimum potential energy is

$\begin{array}{rcl} U (x_{o}) & = & \frac{a}{x_{o}^{12}} - \frac{b}{x_{o}^{6}} \\ = & \frac{a}{{(\frac{2 a}{b})}^{\frac{12}{6}}} - \frac{b}{{(\frac{2 a}{b})}^{\frac{6}{6}}} \\ = & \frac{b^{2} a}{4 a^{2}} - \frac{b^{2}}{2 a} \\ = & - \frac{b^{2}}{4 a} \end{array}$

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

(b)

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

$\frac{d U (x)}{d x} = \frac{- 12 a}{x^{13}} + \frac{6 b}{x^{7}} \frac{d^{2} U (x)}{d x^{2}} = \frac{156 a}{x^{14}} - \frac{42 b}{x^{8}} {\frac{d^{2} U (x)}{d x^{2}}|}_{x_{o}} = \frac{156 a}{{(\frac{2 a}{b})}^{\frac{14}{6}}} - \frac{42 b}{{(\frac{2 a}{b})}^{\frac{8}{6}}} {\frac{d^{2} U (x)}{d x^{2}}|}_{x_{o}} \approx 14 {(\frac{b^{7}}{a^{4}})}^{\frac{1}{3}}$

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) .

$κ = 14 {(\frac{b^{7}}{a^{4}})}^{\frac{1}{3}}$