Griffiths Quantum Mechanics Problem 1.18: Characteristic Size of System

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The discussion centers on the interpretation of intermolecular distance in Griffiths' Quantum Mechanics Problem 1.18, contrasting a spherical model with a cubic box model. The volume per particle is represented differently in each model, leading to a significant factor difference of approximately 1.8. Participants agree that while a box may not fill space completely with spherical particles, it is often used for computational simplicity. The conversation highlights the importance of choosing an appropriate shape for accurate modeling in quantum mechanics. Ultimately, the choice of shape affects the understanding of particle distribution in a given volume.
yucheng
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Homework Statement
The characteristic size (length) of ideal gas where quantum effects are non-negligible is the intermolecular distance, ##d##
Relevant Equations
##pV = Nk_BT##
intermolecular distance means distance between particles. So, I imagine a sphere.

$$\frac{4}{3} \pi d^3 = \frac{V}{N}$$

However, Griffitfhs pictures a box instead, where

$$d^3 = \frac{V}{N}$$

And the difference between both models is a factor of ##(4\pi/3)^{2/5} \approx 1.8##, which is fairly large...
 
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If you fill a box with oranges, does it fill the space completely? Griffiths is correct
 
hutchphd said:
If you fill a box with oranges, does it fill the space completely? Griffiths is correct
I think I understand... So I guess I should have searched for a reliable shape to fill the whole space!
 
And usually we imagine the box to be square, for computational convenience...
 
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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