Work and frequency caused by mutual potential energy

In summary, the mutual potential energy between a Li+ ion and an I- ion can be expressed by the equation U(r)=(-Ke^2/r)+(A/r^10), where K= 9x10^9 N-m^2/C^2 and e=1.6x10^-19 C. The equilibrium distance between the ions is rnot=2.4 Angstrom. To find the work done in eV to separate the ions, A was determined by finding the derivative of U(r) and setting it equal to 0. This value was then used to find the work. For part b, the second derivative of U(r)
  • #1
phys17
2
0
Hi everyone,

Homework Statement


Here is the question:
The mutual potential energy of a Li+ ion and an I- ion as a function of their separation r is expressed fairly well by the equation U(r)=(-Ke^2/r)+(A/r^10), where the first term arises from the Coulomb interaction, and the values of its constant are:

K= 9x10^9 N-m^2/C^2, e=1.6x10^-19 C

The equilibrium distance rnot between the centers of these ions in the LiI molecule is about 2.4 Angstrom. On the basis of this info,

(a) How much work in eV must be done to tear these ions completely away from each other?

(b) Taking the I- ion to be fixed b/c it is so massive, what is the frequency in Hz of the Li+ ion in vibrations of very small amplitude? Take the mass of Li+ as 10^/26



The Attempt at a Solution



I used the U(r) formula to find the work, but how can you find the work without knowing what A is?
For part b, i found the second derivative of the U(r) function to get the value of the spring constant, k. Again, how do we know what the value of A is?

Thanks for any and all help!
 
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  • #2
Hi phys17,

Never in all my years have I encountered a 1/r^10 term in a potential. Am I reading this correctly - you mean 'one over r to the power of ten'? I guess there's nothing actually wrong with that - you can write down any wild potential you want - but I don't think anyone here is going to be able to guess what that term could possibly be or what the 'A' is without some more information. That being said, your ideas on how to deal with the vibrations via a spring constant are very good - well done, I think that is correct! Is it possible that your instructor meant for you to leave things in terms of A? Because with the information interpreted literally and as given, that's the best you can do. Sorry I don't have more insight, but I think we need some more information or your instructor needs to explain what this term is supposed to be.

Hope this helps,
Bill Mills
 
Last edited by a moderator:
  • #3
phys17 said:
Hi everyone,

Homework Statement


Here is the question:
The mutual potential energy of a Li+ ion and an I- ion as a function of their separation r is expressed fairly well by the equation U(r)=(-Ke^2/r)+(A/r^10), where the first term arises from the Coulomb interaction, and the values of its constant are:

K= 9x10^9 N-m^2/C^2, e=1.6x10^-19 C

The equilibrium distance rnot between the centers of these ions in the LiI molecule is about 2.4 Angstrom. On the basis of this info,

(a) How much work in eV must be done to tear these ions completely away from each other?

(b) Taking the I- ion to be fixed b/c it is so massive, what is the frequency in Hz of the Li+ ion in vibrations of very small amplitude? Take the mass of Li+ as 10^/26

The Attempt at a Solution



I used the U(r) formula to find the work, but how can you find the work without knowing what A is?
For part b, i found the second derivative of the U(r) function to get the value of the spring constant, k. Again, how do we know what the value of A is?

Thanks for any and all help!
Welcome to PF :smile:

Here's a hint: what must be true about U(r) at the equilibrium position r=rnot=2.4Å?
billVancouver said:
Hi phys17,

Never in all my years have I encountered a 1/r^10 term in a potential.
I imagine it's an approximation for when then two ions are close together. Somewhat akin to the 1/r12 term in the Lennard-Jones potential typically used for neutral atoms.
Am I reading this correctly - you mean 'one over r to the power of ten'? I guess there's nothing actually wrong with that - you can write down any wild potential you want - but I don't think anyone here is going to be able to guess what that term could possibly be or what the 'A' is without some more information.
Don't forget, we are given the equilibrium position.
That being said, your ideas on how to deal with the vibrations via a spring constant are very good - well done, I think that is correct! Is it possible that your instructor meant for you to leave things in terms of A? Because with the information interpreted literally and as given, that's the best you can do. Sorry I don't have more insight, but I think we need some more information or your instructor needs to explain what this term is supposed to be.

Hope this helps,
Bill Mills
 
  • #4
Thanks so much, guys!

I figured out part a today. To find A, I found the derivative of U(r), set it equal to 0 and then solved for A, since that is the unknown. From there, it's easy to find the work - just substitute all the numbers in.

I ran into trouble for part b. Since the spring constant is the second derivative of U(r), I evaluated at that second derivative. BUT, 2.4 Angstrom converted to meters is a really small value, so when I do A/r^10, I just get 0.
 
  • #5
There are a couple of way around that, that I can think of:

1. Use eV and Å for units, instead of J and m. The numbers should be closer to "regular size" that way, i.e. neither extremely small or extremely large in magnitude.

or

2. Work in scientific notation. Use the calculator to work the between-1-and-10 part of the numbers, and use pen and paper to work the 10xxx's

If you're still stuck, by all means post your work here. Otherwise these suggestions are just a stab in the dark at helping you out.

EDIT added: #3 - use Google as your calculator, it can handle 1/(2.4Å)10:

http://www.google.com/#hl=en&sclien....,cf.osb&fp=3226d37ade666d03&biw=1280&bih=638
 

1. What is work and frequency caused by mutual potential energy?

Work and frequency caused by mutual potential energy refers to the energy that is associated with the interactions between two or more objects. It is the potential energy that is stored in a system due to the position of the objects relative to each other.

2. How is work and frequency caused by mutual potential energy calculated?

Work and frequency caused by mutual potential energy can be calculated using the equation W = ∆PE, where W is the work done on the system and ∆PE is the change in potential energy. Frequency can be calculated using the equation f = 1/T, where T is the period of the oscillation.

3. What is the relationship between work and frequency caused by mutual potential energy?

The relationship between work and frequency caused by mutual potential energy is that as the work done on the system increases, the frequency of the oscillation also increases. This means that the objects are interacting with each other with a greater force, resulting in a higher frequency of oscillation.

4. How does mutual potential energy affect the stability of a system?

Mutual potential energy plays a crucial role in determining the stability of a system. If the mutual potential energy is high, it means that the objects are strongly attracted to each other and the system is stable. On the other hand, if the mutual potential energy is low, the system is less stable and the objects may have a tendency to move apart.

5. What are some real-world examples of mutual potential energy?

There are many real-world examples of mutual potential energy, such as a pendulum swinging, a spring oscillating, or two planets orbiting each other. These examples demonstrate how objects can interact with each other and store potential energy in the form of mutual potential energy.

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