The potential energy function between two atoms in a diatomic molecule can be expressed approximately as follows U ( x ) = x 12 a − x 6 b , where a and b are positive constants and x is the distan

Question

The potential energy function between two atoms in a diatomic molecule can be expressed approximately as follows $U (x) = \frac{a}{x ^{12}} - \frac{b}{x ^{6}}$ , where a and b are positive constants and x is the distance between atoms. What is the dissociation energy of the molecule?

Given, Dissociation energy D= [U(r=∞) - U(at equilibrium)]

TestHub · Accepted Answer

The potential energy function is given by:

$U (x) = \frac{a}{x ^{12}} - \frac{b}{x ^{6}}$

The dissociation energy $D$ is the difference between the potential energy at infinite separation and the potential energy at equilibrium.

1. Find the equilibrium distance $x_{0}$ by minimizing $U (x)$ . This occurs when $\frac{d U}{d x} = 0$ .

$\frac{d U}{d x} = - \frac{12 a}{x ^{13}} + \frac{6 b}{x ^{7}} = 0$

$\frac{6 b}{x ^{7}} = \frac{12 a}{x ^{13}}$

$6 b x^{13} = 12 a x^{7}$

$x^{6} = \frac{12 a}{6 b} = \frac{2 a}{b}$

$x_{0} = (\frac{2 a}{b})^{1/6}$

2. Calculate the potential energy at equilibrium $U (x_{0})$ :

$U (x_{0}) = \frac{a}{x _{0}^{12}} - \frac{b}{x _{0}^{6}} = \frac{a}{( \frac{2 a}{b} ) ^{2}} - \frac{b}{\frac{2 a}{b}} = \frac{a}{\frac{4 a ^{2}}{b ^{2}}} - \frac{b ^{2}}{2 a} = \frac{b ^{2}}{4 a} - \frac{b ^{2}}{2 a} = \frac{b ^{2} - 2 b ^{2}}{4 a} = - \frac{b ^{2}}{4 a}$

3. Calculate the potential energy at infinite separation:

$U (\infty) = 0$

4. Calculate the dissociation energy $D$ :

$D = U (\infty) - U (x_{0}) = 0 - (- \frac{b ^{2}}{4 a}) = \frac{b ^{2}}{4 a}$

Therefore, the dissociation energy of the molecule is $\frac{b ^{2}}{4 a}$ .

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