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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

The Fermi velocity $V_{f}$ is defined by $E_{F} = \frac{1}{2} m s_{F}^{2}$ , where $E_{F}$ is the Fermi energy. The Fermi energy for conduction electron in sodium is 3 I $V$ . (a) Calculate the Fermis velocity (b) What would be the wavelength of an electron with this velocity? (c) If each sodium atom contributes one conduction electron to the electron gas and sodium atom are spaced roughly $0 .37 nm$ apart. If it is necessary, by the criteria of equation (9-43), to treat the conduction electron gas as quantum gas?

A particle subject to a planet's gravitational pull has a total mechanical energy given by $E_{mechanical} = \frac{1}{2} m v^{2} - \frac{G M m}{r}$ , whereis the particle's mass.M the planet's mass, and Gthe gravitational constant $6.67 \times 10^{- 11}$ $N \cdot m^{3} / {kg}^{2}$ . It may escape if its energy is zero that is, if its positive KE is equal in magnitude to the negative PE holding if to the surface. Suppose the particle is a gas molecule in an atmosphere.

(a) Temperatures in Earth's atmosphere may reach $1000 K$ . Referring to the values obtained in Exercise 45 and given that $R_{Earth} = 6.37 \times$ $10^{6} m$ and $M_{Earth} = 5.98 \times 10^{24} kg$ . should Earth be able to "hold on" to hydrogen $(1 g / mol)$ ? 10 nitrogens $(28 g / mol)$ ? (Note: An upper limit on the number of molecules in Earth's atmosphere is about $10^{{}^{- 18}}$ ).

(b) The moon's mass is $0.0123$ times Earth's. its radius 0.26 times Earth's, and its surface temperatures rise to $370 K$ . Should it be able to hold on to these gases?

Four distinguishable Hamonic oscillators $a, b, c, and d$ may exchange energy. The energies allowed particleare $E_{a} = n_{a} ℏ ω_{0}$ : those allowed particlebare $E_{b} = n_{b} ℏ ω_{0}$ , and so on. Consider an overall state (macro-state) in which the total energy is $3 ℏ ω_{0}$ . One possible microstate would have particles $a, b, and c$ in their $n = 0$ states and particle d in its $n = 3$ states that is, $(n_{a}, n_{b}, n_{c} n_{d}) = (0, 0, 0, 3)$ .

(a) List all possible microstates, (b) What is the probability that a given particle will be in its $n = 0$ state? (c) Answer part (b) for all other possible values of n. (d) Plot the probability versus n.

According to Wien's law, the wavelength $λ_{m a x}$ at which the thermal emission of electromagnetic energy from a body of temperature $T$ is maximum obeys $λ_{m a x} T = 2 .898 \times 10^{- 3} m \cdot K$ .Show that this law follows from equation (9-47). To do this. Use $f = c / λ$ to expressin terms of $λ$ rather than f, then obtain an expression that, when solved, would yield the wavelength at which this function is maximum. The transcendental equation cannot be solved exactly, so it is enough to show that $λ = (2 .898 \times 10^{- 3} m \cdot K) / T$ solves it to a reasonable degree of precision.

Suppose that in Figure 9.27, the level labelled $E_{1}$ , rather than the one labelled $E_{2}$ , were metastable. Might the material still function as a laser? Explain.

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