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Chapter 9: Q58E (page 407)

Show that. using equation $(9 - 36)$ , density of states $(9 - 38)$ follows fromlocalid="1658380849671" $(9 - 37)$

Short Answer

Expert verified

The expression for density states is $\frac{m^{\frac{3}{2}} L^{3}}{π^{2} ℏ^{3} \sqrt{2}} E^{\frac{1}{2}}$ .

Step by step solution

01

Formula used

The expression for density of energy state is given by,

$D (E) = \frac{Differentialnumberofstatesinrange d E}{d E}$

The expression for radial distance from origin is given by,

$n = \sqrt{\frac{2 m L^{2} E}{π^{2} ℏ^{2}}}$

02

Expression for density of energy state

The expression for density of energy state is calculated as,

$\begin{matrix} D (E) = \frac{Differentialnumberofstatesinrange d E}{d E} \\ = \frac{\frac{1}{8} 4 π n^{2} d n}{d E} \\ = \frac{1}{8} 4 π {(\sqrt{\frac{2 m L^{2} E}{π^{2} ℏ^{2}}})}^{2} \frac{d}{d E} (\sqrt{\frac{2 m L^{2} E}{π^{2} ℏ^{2}}}) \\ = (\frac{π}{2}) (\frac{2 m L^{2} E}{π^{2} ℏ^{2}}) (\frac{L}{2 π ℏ}) \sqrt{\frac{2 m}{E}} \end{matrix}$

On further solving,

$\begin{matrix} D (E) = \frac{L^{3}}{π^{2} ℏ^{3}} \sqrt{\frac{2 m^{3} E^{2}}{4 E}} \\ = \frac{L^{3}}{π^{2} ℏ^{3}} \sqrt{\frac{m^{3} E}{2}} \\ = \frac{m^{\frac{3}{2}} L^{3}}{π^{2} ℏ^{3} \sqrt{2}} E^{\frac{1}{2}} \end{matrix}$

Therefore, the expression for density states is . $\frac{m^{\frac{3}{2}} L^{3}}{π^{2} ℏ^{3} \sqrt{2}} E^{\frac{1}{2}}$

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Most popular questions from this chapter

In a large system of distinguishable harmonic oscillator how high does the temperature have to be for the probable number of particles occupying the ground state to be less than 1 ?

Consider a room divided by imaginary lines into three equal parts. Sketch a two-axis plot of the number of ways of arranging particles versus $N_{left}$ and $N_{right}$ for the case $N = 10^{23}$ , Note that $N_{middle}$ is not independent, being of course $N - N_{nght} - N_{left}$ Your axes should berole="math" localid="1658331658925" $N_{left}$ and $N_{right}$ , and the number of ways should be represented by density of shading. (A form for numbers of ways applicable to a three-sided room is given in Appendix I. but the question can be answered without it.)

At high temperature, the average energy of a classical one-dimensional oscillator is $k_{B} T$ , and for an atom in a monatomic ideal gas. it is $\frac{1}{2} k_{B} T$ . Explain the difference. using the equipartition theorem.

Figure 9.8 cannot do justice to values at the very highspeed end of the plot. This exercise investigates how small it really gets. However, although integrating the Maxwell speed distribution over the full range of speeds from 0 to infinity can be carried out (the so-called Gaussian integrals of Appendix K), over any restricted range, it is one of those integrals that. unfortunately. cannot be done in closed form. Using a computational aid of your choice. show that the fraction of molecules moving faster thanis; faster than $6 v_{rms}$ is $- 10^{- 23}$ ; and faster than $10 v_{ms}$ is $~ 10^{- 64}$ . where $v_{r m s}$ " from Exercise 41, is $\sqrt{3 k_{B} T / m}$ . (Exercise 48 uses these values in an interesting application.)

Not surprisingly. in a collection of oscillators, as in other thermodynamic systems, raising the temperature causes particles' energies to increase. Why shouldn’t point be reached where there are more panicles in some high energy state than in a lower energy. state? (The fundamental idea, not a formula that might arise from it. is the object.)

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