The potential energy of a diatomic molecule in terms of interatomic distance R is given by $$ U(R)=-\frac{a}{R^{\mathrm{m}}}+\frac{b}{R^{n}} $$ ia) Derive an expression for the equilibrium spacing of the atoms and hence obtain the dissociation (b, Explain die impose of equilibrium spacing as die configure, ion of lower, polcn.inl energy and hence obtain an expression for the dissociation energy of the atoms (c ) Prove n > m for U( R ) to be minimum at R = Rc

To find the equilibrium spacing of atoms in a diatomic molecule, you need to determine where the potential energy is at its lowest. The equilibrium spacing, denoted as \(R_c\), is the distance between the two atoms where this minimum occurs.

The potential energy function given is \[U(R) = -\frac{a}{R^m} + \frac{b}{R^n}\], where \(a\) and \(b\) are constants, \(R\) is the interatomic distance, \(m\) relates to the attractive force, and \(n\) relates to the repulsive force. Turning points, including where the minimum energy lies, can be found by taking the first derivative of \(U(R)\) with respect to \(R\) and setting it to zero.

The critical point, or equilibrium spacing, is given by \[R_c = \left( \frac{b \cdot n}{a \cdot m} \right)^\frac{1}{n-m}\]. This equation signifies that the drop in potential energy to the equilibrium state involves a balance between attractive and repulsive forces.

Dissociation energy is the energy required to break a molecule into its individual atoms. For a diatomic molecule, it represents the energy needed to separate the atoms from their equilibrium distance \(R_c\) to an infinite distance.

At the equilibrium spacing \(R_c\), the potential energy is \[U(R_c) = -\frac{a}{R_c^m} + \frac{b}{R_c^n}\]. As the atoms separate infinitely, the potential energy, \(U(\infty)\), approaches zero. The dissociation energy \(E_d\) can then be defined as the energy difference needed to move from \(R_c\) to infinity: \[E_d = -U(R_c) = \frac{a}{R_c^m} - \frac{b}{R_c^n}\].

This calculation emphasizes that the dissociation energy depends inversely on the equilibrium distance raised to the powers of \(m\) and \(n\).

To confirm that \(R_c\) is indeed a minimum distance, we apply the second derivative test. After finding the first derivative and setting it to zero to locate potential equilibrium points, the second derivative is used to determine the nature of these points - whether they are minima, maxima, or inflection points.

The second derivative of the potential energy function \(U(R)\) is: \[ \frac{d^2U}{dR^2} = -\frac{a \cdot m \cdot (m+1)}{R^{m+2}} + \frac{b \cdot n \cdot (n+1)}{R^{n+2}} \]. For \(R_c\) to be a minimum, the second derivative at \(R_c\) should be positive. Substituting \(R_c\) into this second derivative and ensuring the result is positive confirms this.

This positive second derivative verifies the equilibrium is stable and at its minimum potential energy, which aligns with physical expectations for a stable molecular structure.

In diatomic molecules, the balance between attractive and repulsive forces determines the potential energy curve and the stability of the molecule. The function provided, \[U(R) = -\frac{a}{R^m} + \frac{b}{R^n}\], showcases the interplay between these forces.

Attractive Forces: Represented by the negative term \(-\frac{a}{R^m}\), they dominate at larger distances, causing the atoms to pull toward each other.

Repulsive Forces: Represented by the positive term \(+\frac{b}{R^n}\), they dominate when atoms are very close, pushing them apart to prevent overlap of electronic clouds.

A stable molecule is formed when these opposing forces equilibrate at \(R_c\), resulting in the least potential energy. Physical intuition dictates that for a minimum potential energy point, where the molecule is most stable, the repulsive forces must increase more quickly than the attractive as \(R\) decreases. This is mathematically represented by \(n > m\), ensuring the necessary equilibrium and stability.