Consider N distinguishable 3-dimensional harmonic oscillators of HamiltonianH(q, p) = P mw²q?2m23Ni=1This is a simple model of the vibrations of atoms in a solid.1. Calculate the umber of accessible states 2 for a given energy E.Since N is very large, you may assume that the number of state of energy E is thesame as the number of states of energy

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Consider N distinguishable 3-dimensional harmonic oscillators of Hamiltonian
H(q, p) = P mw²q?
2m
2
3N
i=1
This is a simple model of the vibrations of atoms in a solid.
1. Calculate the umber of accessible states 2 for a given energy E.
Since N is very large, you may assume that the number of state of energy E is the
same as the number of states of energy <E
2. Calculate the entropy S, the temperature T, the energy E and the heat capacity C.

Expert Solution

Step 1 Concept

(1) Calculate the number of accessible states for the given energy.

Solution:

If the solid crystal is made up of N atoms, the motion of each of them has three independent components. An atom can not move freely but can vibrate about its equilibrium position. That means vibrations of an atom are equivalent to vibration of three harmonic oscillators.

As given, N distinguishable 3-dimensional harmonic oscillator of Hamiltonian is given by;

$\begin{array}{rcl} H (q, p) & = & \sum_{i = 1}^{3 N} [\frac{{p_{i}}^{2}}{2 m} + \frac{m ω^{2} {q_{i}}^{2}}{2}] \\ T h i s c a n b e w r i t t e n i n g e n e r a l f o r m a s; \\ H & = & \frac{p^{2}}{2 m} + \frac{m ω^{2} x^{2}}{2} \\ B u t H & = & E \\ ∴ E & = & \frac{p^{2}}{2 m} + \frac{1}{2} k x^{2} \\ T h e r e f o r e; \\ <H> & = & E_{n} = (n + \frac{1}{2}) ħ ω \\ w h e r e; n & = & 1.2.3 . . . . . . \infty \\ T h e p a r t i t i o n f u n c t i o n i s g i v e n b y; \\ Z & = & \sum_{n} e^{- β E_{n}} \\ Z & = & \sum_{n = 1}^{\infty} e^{- β (n + \frac{1}{2}) ħ ω} \\ Z & = & \sum_{n = 1}^{\infty} e^{- β n ħ ω} \times e^{\frac{^{- β ħ ω}}{2}} \\ Z & = & e^{\frac{- β ħ ω}{2}} \sum_{n = 1}^{\infty} e^{- β n ħ ω} \\ Z & = & e^{\frac{- β ħ ω}{2}} [1 + e^{- β ħ ω} + e^{- 2 β ħ ω} + e^{- 3 β ħ ω} + . . . . . . . . . .] \\ B u t, [1 + e^{- β ħ ω} + e^{- 2 β ħ ω} + e^{- 3 β ħ ω} + . . . . . . . . . .] & = & g e o m e t r i c p r o g r e s s i o n w h i c h g i v e s \\ [1 + e^{- β ħ ω} + e^{- 2 β ħ ω} + e^{- 3 β ħ ω} + . . . . . . . . . .] & = & \frac{1}{(1 - e^{^{- β ħ ω}})} \\ T h e r e f o r e; \\ Z & = & \frac{e^{\frac{- β ħ ω}{2}}}{(1 - e^{^{- β ħ ω}})} \\ T h i s i s f o r a n a t o m w i t h 3 h a r m o n i c o s c i l l a t o r s \\ F o r N a t o m s 3 N h a r m o n i c o s c i l l a t o s a r e r e q u i r e d . \\ Z_{3 N} & = & {[Z_{1}]}^{3 N} \\ Z_{3 N} & = & {[\frac{e^{\frac{- β ħ ω}{2}}}{(1 - e^{^{- β ħ ω}})}]}^{3 N} \\ Z_{3 N} & = & {[\frac{1}{(e^{\frac{^{+ β ħ ω}}{2}} - e^{\frac{^{- β ħ ω}}{2}})}]}^{3 N} \\ The number of accecible states is given by \\ Ω & = & \frac{V^{N}}{(\frac{3 N}{2})!} {(\frac{2 {πmE}_{n}}{h^{2}})}^{\frac{3 N}{2}} \\ Ω & = & \frac{V^{N}}{(\frac{3 N}{2})!} {(\frac{2 πm (n + \frac{1}{2}) ħω}{h^{2}})}^{\frac{3 N}{2}} \\ Answer & = & The number of accesible statesfor the given energy is; \\ Ω & = & \frac{V^{N}}{(\frac{3 N}{2})!} {[\frac{2 πm (n + \frac{1}{2}) ħω}{h^{2}}]}^{\frac{3 N}{2}} \end{array}$

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