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CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=\left(a / r^{12}\right)-\left(b / r^{6}\right),\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs: one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\). (b) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the cquilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{~m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{~J}\) per molecule. Find the values of the constants \(a\) and \(b\).

Short Answer

Expert verified
Force \(F(r) = 12\frac{a}{r^{13}} - 6\frac{a}{r^{7}}\), equilibrium distance \(r_{eq} = \sqrt[6]{2a / b}\), dissociation energy \(E = -min(U(r))\). Constants \(a\) and \(b\) can be determined using known data for specific molecules.

Step by step solution

01

Calculate the Force \(F(r)\)

The force between the atoms can be calculated using \(F(r) = -\frac{dU}{dr}\). Taking derivative of \(U(r)\) with respect to \(r\), we obtain \(F(r) = 12\frac{a}{r^{13}} - 6\frac{a}{r^{7}}\)
02

Graph \(U(r)\) and \(F(r)\)

According to the functions \(U(r)\) and \(F(r)\), their plots could be sketched. Since these calculations are complex and are not easily represented using text, this step would often involve learning how to plot functions on a graphing calculator or mathematics software.
03

Determine the Equilibrium Distance

Equilibrium occurs when force \(F(r) = 0\). Setting our equation for \(F(r)\) equal to zero, we can solve for \(r\) by equating \(12a / r^{13} = 6b / r^{7}\). Hence, \(r_{eq} = \sqrt[6]{2a / b}\). The equilibrium is stable if \(d^2U / dr^2 > 0\), ie. the second derivative of \(U(r)\) at \(r = r_{eq}\) is greater than zero.
04

Calculate the Dissociation Energy

The dissociation energy can be found by setting \(r\) to infinity in \(U(r)\). In practice, we can calculate the potential energy at the point of equilibrium and find the difference with \(U\) at infinity. Therefore, the dissociation energy \(E\) can be found using \(E = U(r -> infinity) - U(r_{eq}) = -min(U(r))\).
05

Determine the Values of \(a\) and \(b\)

Using the known equilibrium distance and the dissociation energy for the molecule CO, the values of \(a\) and \(b\) can be found using the equations obtained in steps 3 and 4. This may involve solving simultaneous equations.

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Key Concepts

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

Potential Energy of Diatomic Molecule
Potential energy in a diatomic molecule offers valuable insight into how the two atoms within the molecule interact and stay bonded. The potential energy, represented as U(r), changes depending on the distance r between the atoms. The function U(r) = (a / r^{12}) - (b / r^{6}), known as the Lennard-Jones potential, illustrates this relationship graphically.

At far distances, the interaction is weak and the energy is high due to repulsion, modeled by the r^{12} term. As the atoms come closer to each other, the energy decreases reaching a minimum, indicating a strong attraction, captured by the -r^{6} term. When plotted, the graph shows a curve with a well-defined minimum that corresponds to the most stable configuration of the two atoms, with the least potential energy.

Visualizing this function can help students understand why molecules have certain stable distances at which the energy is minimized, thus laying the groundwork for the study of chemical bonds and molecular structures.
Equilibrium Distance in Molecular Physics
Understanding equilibrium distance in molecular physics is fundamental in explaining how atoms settle into a stable configuration. The equilibrium distance, r_{eq}, is the specific separation at which the force exerted on each atom by the other is zero. Mathematically, we find this by setting the first derivative of the potential energy, U(r), to zero, since the force F(r) is the negative gradient of U(r).

Thus, at the equilibrium distance, the attractive and repulsive forces between the atoms balance each other out. For the Lennard-Jones potential, the equilibrium distance can be expressed as r_{eq} = \(\sqrt[6]{{2a / b}}\). This concept is pivotal in understanding molecular sizes and bond lengths in chemistry and physics. It also helps in rationalizing the stable geometries of molecules.
Interatomic Forces and Stability
Interatomic forces are the result of interactions between charged particles within atoms. These forces govern the stability of diatomic molecules. Stability at the equilibrium distance can be determined by evaluating the curvature of the potential energy graph around r_{eq}. If the second derivative of U(r) at this point is positive, the molecule is stable. This indicates that any small displacement of the atoms would result in a restoring force pushing them back towards equilibrium.

Exploring Stability

When delving into the concept of stability, it's essential to understand that a stable equilibrium is like a valley on the energy graph, whereas an unstable one is like a hilltop. The concepts of minima and maxima in the energy landscape are fundamental in predicting how molecules will behave when subjected to disturbances.

Considering the dissociation energy is also a significant part of analyzing molecular stability. It's the minimum energy required to break the bond between atoms at the equilibrium distance, effectively pushing the atoms over the energy 'hill' to an infinite separation. The larger the dissociation energy, the more stable the molecule. These principles are crucial in fields such as material science, where the stability of compounds under various conditions is key.

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Most popular questions from this chapter

A wooden block with mass \(1.50 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of \(6.00 \mathrm{~m}\) up the incline from \(A\), the block is moving up the incline at \(7.00 \mathrm{~m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.50\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

BIO Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a \(250 \mathrm{~g}\) object was hung from it, the tendon stretched \(1.23 \mathrm{~cm}\). (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\). (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing. and how much energy is stored in it at that point?

A 25.0 kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a 65 kg hiker eats one bar, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that \(100 \%\) of the food calories that are eaten are absorbed and used by the body. This is not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the body eliminates the rest. Metabolic efficiency varies considerably from person to person.

A system of two paint buckets connected by a lightweight rope is released from rest with the \(12.0 \mathrm{~kg}\) bucket \(2.00 \mathrm{~m}\) above the floor (Fig. \(\mathbf{P 7 . 5 1}\) ). Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. Ignore friction and the mass of the pulley.

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