Four lowest vibrational states of these ions

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SUMMARY

The discussion focuses on calculating the energies of the four lowest vibrational states of aluminum ions in a crystal lattice with an equilibrium spacing of 0.30 nm. The participant initially used the formula for equilibrium separation, \( \sqrt{\hbar/m\omega} \), to derive the angular frequency \( \omega \) and applied the vibrational energy formula \( (n-1)\hbar \omega \). However, the calculated energies of 0.42 eV, 1.30 eV, 2.1 eV, and 2.9 eV were incorrect, indicating a misunderstanding in the application of the energy level formula and the derivation of frequency.

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Homework Statement


41.CP44.jpg

In most metals, the atomic ions form a regular arrangement called a crystal lattice. The conduction electrons in the sea of electrons move through this lattice. The figure (Intro 1 figure) is a one-dimensional model of a crystal lattice. The ions have mass m , charge e , and an equilibrium separation b.


Homework Equations


Suppose this crystal consists of aluminum ions with an equilibrium spacing of 0.30nm. What are the energies of the four lowest vibrational states of these ions?


The Attempt at a Solution


well i thought separation equilibrium = (h bar/mw)^0.5 from this i got the w angular frequency and then i use the vibrational formula (n-1)h bar * w i got these first four vibrational states 0.42ev,1.30ev,2,1ev,2,9ev but i got it wrong can anybody tell what i did wrong please thanks
 
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You'll have to give details as to how you derived that frequency. Also, there is a problem with the formula for the energy levels.
 

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