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Consider a hypothetical relation giving potential energy of a system of two atoms in a diatomic molecule: \(U-\frac{\alpha}{r^{11}} \frac{\beta}{r^{5}}\), (where \(\alpha\) and \(\beta\) are constant, \(r\) represents interatomic scparation). Interatomic separation at the equilibrium is (a) \(\left(\frac{11 \alpha}{5 \beta}\right)^{1 / 6}\) (b) \(\left(\frac{11 \alpha}{\beta}\right)^{1 / 6}\) (c) \(\left(\frac{5 \beta}{11 \alpha}\right)^{1 / 6}\) (d) \(\left(\frac{\beta}{11 \alpha}\right)^{1 / 6}\)

Short Answer

Expert verified
The correct answer is (a) \( \left(\frac{11\alpha}{5\beta}\right)^{1/6} \).

Step by step solution

01

Understanding the Potential Energy Equation

The given potential energy is represented as \( U = -\frac{\alpha}{r^{11}} - \frac{\beta}{r^{5}} \). Our goal is to find the value of \( r \) at equilibrium, where the potential energy is at a minimum.
02

Setting the First Derivative to Zero

To find the equilibrium separation, set the first derivative of the potential energy equation with respect to \( r \) equal to zero: \( \frac{dU}{dr} = 0 \). This corresponds to finding the minimum of the function.
03

Deriving the Derivative

Calculate the derivative: \[ \frac{dU}{dr} = \frac{d}{dr}\left(-\frac{\alpha}{r^{11}} - \frac{\beta}{r^{5}}\right) = 11\frac{\alpha}{r^{12}} + 5\frac{\beta}{r^{6}} \].
04

Solving for r

Set the derivative equal to zero to solve for \( r \): \( 11\frac{\alpha}{r^{12}} + 5\frac{\beta}{r^{6}} = 0 \). Rearrange this to \( 11\alpha r^{6} + 5\beta r^{12} = 0 \). Remove common terms and solve for \( r \): \( 11\alpha = -5\beta r^{6} \).
05

Simplifying the Equation

Rearranging gives \( r^{6} = \frac{11\alpha}{5\beta} \). Thus, \( r = \left(\frac{11\alpha}{5\beta}\right)^{1/6} \).
06

Selecting the Correct Option

Review the options provided and match them with our solution. The correct match is option (a), \( \left(\frac{11\alpha}{5\beta}\right)^{1/6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Separation
In the world of diatomic molecules, equilibrium separation refers to the specific distance between two atoms where the potential energy is at its lowest. This minimal potential energy state is often desired because it ensures that the atoms are as stable as possible, aligning with the concept of equilibrium in physics.

To find this equilibrium separation, we need to identify the point where the forces acting on the atoms are balanced. Mathematically, we achieve this by differentiating the potential energy function with respect to interatomic distance \( r \), and then setting this derivative to zero to find the minimum potential energy. This process leads us to the formula for equilibrium separation: \( r = \left(\frac{11\alpha}{5\beta}\right)^{1/6} \), demonstrating the balance of attractive and repulsive forces in the molecule. By achieving equilibrium, the system remains in a stable state.
Diatomic Molecule
Diatomic molecules are among the simplest types of molecules, composed of only two atoms. These atoms can be either the same element, like oxygen (O\(_2\)) or different elements, such as hydrogen chloride (HCl).

The behavior and properties of diatomic molecules are heavily influenced by the intermolecular forces between the atoms. They are bound together through chemical bonds, involving electron sharing in covalent bonds or electron transfer in ionic bonds. Understanding the energy landscape, including potential energy, is crucial in predicting how these molecules behave. The interplay of forces within a diatomic molecule determines its structural stability and reactivity, especially at equilibrium separation where potential energy is minimized.

These molecules are foundational to studies in chemistry and physics, illustrating the basic principles of molecular dynamics and interactions.
Interatomic Forces
Interatomic forces are essential to understanding the structure and interactions of atoms within molecules. These forces can be broadly classified into two categories:
  • **Attractive Forces**: These include forces such as van der Waals forces and covalent bonding, which hold atoms together.
  • **Repulsive Forces**: These forces act to push atoms apart when they get too close, avoiding overlap of electron clouds.
The balance between these forces dictates the arrangement of atoms in a molecule. When attractive and repulsive forces are equal, the potential energy attains a minimum value, corresponding to the equilibrium separation.

In the context of a potential energy equation like \( U = -\frac{\alpha}{r^{11}} - \frac{\beta}{r^{5}} \), the term \( -\frac{\beta}{r^{5}} \) relates to shorter-range attractive forces, whereas \( -\frac{\alpha}{r^{11}} \) reflects longer-range repulsive forces. Analyzing changes in interatomic forces helps in understanding not just equilibrium conditions but also how atoms respond under different energy states, vital in designing and manipulating materials at the molecular level.

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Most popular questions from this chapter

In a molecule, the potential energy between two atoms is given by: \(U(x)-\frac{a}{x^{12}}-\frac{b}{x^{6}}\), where \(a\) and \(b\) are positive constants and \(x\) is the distance between atoms. For equilibrium of atom, the valuc of \(x\) is (a) zero (b) \(\left[\frac{a}{2 b}\right]^{1 / 6}\) (c) \(\left[\frac{2 a}{b}\right]^{1 / 6}\) (d) \(\left[\frac{11 a}{5 b}\right]^{1 / 6}\)

A particle of mass \(m\) (starting at rest) moves vertically upwards from the surface of earth under an external force \(\vec{F}\) which varies with height \(z\) as \(\vec{F}-(2-\alpha z) m \vec{g}\), where \(\alpha\) is a positive constant. If \(H\) is the maximum height to which particle rises. Then (a) \(H-\frac{1}{\alpha}\) (b) Work done by \(\vec{F}\) during motion upto \(\frac{H}{2}\) is \(\frac{3 m g}{2 \alpha}\) (c) \(H-\frac{2}{\alpha}\) (d) Velocity of particle at \(\frac{H}{2}\) is \(\sqrt{\frac{g}{\alpha}}\)

Potential energy of a particle moving along \(x\) -axis is given by \(U=(a x-b) x\). Speed of the particle is maximum at \(x\) equal to (here \(a\) and \(b\) are positive constants) (a) Zero (b) \(\frac{b}{2 a}\) (c) \(\frac{b}{a}\) (d) \(\frac{2 b}{a}\)

A force \(F--\frac{k}{x^{2}}(x \neq 0)\) acts on a particle in \(x\) -direction. Find the work done by this force in displacing the particle from \(x=a\) to \(x=+2 a .1\) lere \(k\) is a positive constant.

\(\Lambda\) body is projected with velocily \(v\) and angle of projection is \(\theta\) Column-I (a) Variation of kinetic energ \(y(K)\) with respect to horizontal position \((x)\) is (b) Variation of poetntial cnergy \((U)\) with respect to horizontal position \((x)\) is (c) Kinetic energy at the highest point is Column-II (p) Lincar (q) Parabolic (r) I Iyperbolic (s) Maximum (l) Minimum

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