Hello,
I don’t understand how I can derive the distance where the potential becomes anharmonic at the Lennard Jones model of question 2.5
Kind regards,
Fenna
I will discuss this at today’s Q&A
Conceptually this is one of the harder questions in the course, even though there isn’t a lot of calculations involved. It also applies a very important way of reasoning in physics.
So we are looking at the Lennard-Jones potential
U = 4 \epsilon\left[\left(\frac{r}{\sigma}\right)^{12} - \left(\frac{r}{\sigma}\right)^6\right].
We want to know how many phonons fit into this potential before it becomes anharmonic.
The potential has 2 parameters: \epsilon that tells the magnitude of energies that are involved, and \sigma, that controls over which length the potential changes. As long as we’re asking questions about only this potential, our answers cannot depend on anything else. We know that a minimum of the potential exists because at small r the first term takes over, and at large r the second one. Let’s make a list of estimates that we can perform only with the observations above:
- Q: What is the minimum of the potential energy?
A: U_0 \sim - \epsilon - Q: At what r_0 is the potential energy minimal?
A: r_0 \sim \sigma - Q: How large should be \delta r so that \delta U\sim \epsilon?
A: \delta r \sim \sigma - Q: What is the elastic constant \kappa?
A: \kappa \sim \epsilon / \sigma^2 (there’s only one way to get an answer with the correct dimension). - Q: What is the coefficient \kappa_3 of the third order expansion term around r_0?
A: \kappa_3 \sim \epsilon / \sigma^3 - Q: At what distance does the quadratic term become comparable with qubic?
A: At \delta r \sim \sigma - Q: At what energy above the minimum does the potential become anharmonic?
A: At \delta U \sim \epsilon
The last observation is important, but we also need to know the number of phonons. Phonons are related to the device dynamics, and we cannot argue about those based on the shape of the potential alone. Thankfully, the remaining step doesn’t take much. We know that the phonon frequency is \omega \sim \sqrt{\kappa / m} and that energy of a system with n phonons is \sim n \hbar \omega. Comparing this to the answer of question 7 above we get n\hbar\omega \sim \epsilon \Rightarrow n \sim \epsilon / \hbar \sqrt{\kappa / m } = \sigma \sqrt{m\epsilon} / \hbar .
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Hello, just to follow up no this question because I feel it is still unclear how the expression for the r_anharmonic is arrived at. Additionally the k3 term is proportional to 1/sigma^3 around r0. I tried equating the terms and attempting to find a relation but couldn’t, could a bit more explanation be given please?
Thank you!
You’re right, and I had a typo (corrected now).
To get r_\text{anharmonic} we use
\kappa r_\text{anharmonic}^2 \sim \kappa_3 r_\text{anharmonic}^3,
basically saying that the two terms become comparable at that distance.
Why is the energy required to break the molecule apart equal to the ground state energy of the molecule (but negative)? (Question 2.4)
This is how I understand it:
When a molecule is broken, the atoms of the atom have no covalent bond anymore. This means that no energy can be stored in a covalent bond between those atoms.
When an energy E is stored in the covalent bonds of a molecule, you have to remove all that energy of the covalent bonds to break that atom. Therefore, the amount of energy required to break the molecule is -E.
If this explanation is true, the amount of energy to break the molecule can be larger then the ground state energy when the covalent bond(s) are in an excited energy level. Is this true? I am doubting because this is not mentioned in the answer model.
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Your explanation is correct, but do you mean that the required energy can be smaller if the molecule is in an excited state?
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No, if I understand you well, I do not mean that.
I meant that the required energy equals to the total energy stored in the covalent bonds in a molecule. All the energy. Therefore, when those bonds are in an excited state, the energy required to break the molecule is larger then the ground state energy which is mentioned in the answer.
Because in the question was not given that the covalent bond was in the ground state, I thought we had to give also an expression of the possible energy which should be added due to the the possible excited states.
Ps. Thank you for your fast response.
Pss. Tiny error: in your first post in this thread the 12 and 6 in the Lennard-Jones potential should be -12 and -6.