Quantized sound waves that propagate in a crystal lattice are called phonons. It energy is e = ho = huk, where u is the speed of sound and k is the usual wavenumber. The density of states is given by 3V €² 2л²h³v³ g(e)de = According to Debye, this value for the density of states is true up to some ceiling value €p, called the Debye energy. The density of states then vanishes at values € > €D. a. Solve for the Debye energy if the density of states satisfy the following Hint: Change the bounds of integration. -de [.. g(e)de = 3N.
Quantized sound waves that propagate in a crystal lattice are called phonons. It energy is e = ho = huk, where u is the speed of sound and k is the usual wavenumber. The density of states is given by 3V €² 2л²h³v³ g(e)de = According to Debye, this value for the density of states is true up to some ceiling value €p, called the Debye energy. The density of states then vanishes at values € > €D. a. Solve for the Debye energy if the density of states satisfy the following Hint: Change the bounds of integration. -de [.. g(e)de = 3N.
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Transcribed Image Text:Quantized sound waves that propagate in a crystal lattice are called phonons. It energy is
e = ho = huk,
where u is the speed of sound and k is the usual wavenumber. The density of states is given by
3V €²
2л²h³v³
g(e)de =
According to Debye, this value for the density of states is true up to some ceiling value €p, called the Debye
energy. The density of states then vanishes at values € > €D.
a. Solve for the Debye energy if the density of states satisfy the following
Hint: Change the bounds of integration.
-de
[..
g(e)de = 3N.
Transcribed Image Text:b. Use the Debye energy to show that the density of states can be written as
for 0 <e < ep, then zero otherwise.
g(e)de = 9N!
-de
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