Calculating Ion Distance in a 1D Lattice: A Simple Solution

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To calculate ion distance in a 1D lattice, it's essential to clarify the problem by providing an attempted solution. The discussion emphasizes the importance of considering only nearest neighbor interactions due to the rapid drop-off of repulsive potential compared to attractive forces. It suggests referencing introductory solid-state physics texts, such as Kittel's, for foundational concepts. Additionally, the dimensionality of the lattice should be specified, with a recommendation to start with the simplest case. Clear definitions and assumptions are crucial for accurate calculations.
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Homework Statement
Calculate the distance between the nearest ions at which the two forms of the repulsive potential give equal lattice energies.
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"nearest ions" suggests there are more than two in the scenario. Is there more intro to this question?
 
In the simplest models of lattice, the repulsive potential is considered to drop much faster than the attractive one so the repulsive effects are considered only between the nearest neighbors in the calculations of the total energy. This is how it is done is some introductory solid-state books, like Kittel's.
 
Also are we to assume a 1 dimensional lattice? 2D? 3D? Choose the simplest case and attempt a solution.....
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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