Chapter 9: Q30E (page 404)

Obtain equation (9- 15) from (9-14). Make use or the following sums, correct when $| x | < 1$ :
$\begin{array}{l} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \\ \sum_{n = 0}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}} \end{array}$

Short Answer

Expert verified

Equation $\bar{E} = \frac{\sum_{n = 0}^{\infty} n h ω_{0} e^{- \frac{m ω_{0}}{k_{B} T}}}{\sum_{n = 0}^{\infty} e^{\frac{m h ω_{0}}{k_{B} T}}}$ simplifies to equation $\bar{E} = \frac{h ω_{0}}{e^{\frac{h_{ω_{0}}}{k_{B} T}} - 1}$

Step by step solution

Average Energy.

Let us consider the expression for the average energy:

$\bar{E} = \frac{\sum_{n = 0}^{\infty} n h ω_{0} e^{\frac{- n h ω_{0}}{k_{B} T}}}{\sum_{n = 0}^{\infty} e^{\frac{- n h ω_{0}}{k_{B} T}}}$

Performing some minor modifications, we have:

$\bar{E} = h ω_{0} \frac{\sum_{n = 0}^{\infty} n e^{\frac{- n h ω_{0}}{k_{B} T}}}{\sum_{n = 0}^{\infty} e^{\frac{- n h ω_{0}}{k_{B} T}}}$

$\bar{E} = \frac{\sum_{n = 0}^{\infty} n h ω_{0} e^{\frac{- n h ω_{0}}{k_{B} T}}}{\sum_{n = 0}^{\infty} e^{\frac{- n h ω_{0}}{k_{B} T}}}$

Properties

Making use of the property $x = e^{\frac{- h ω_{0}}{k_{B} T}}$ .

$\begin{array}{l} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \\ \Rightarrow \sum_{n = 0}^{\infty} {[e^{\frac{- h ω_{0}}{k_{B} T}}]}^{n} = \frac{1}{1 - e^{\frac{- h ω_{0}}{k_{B} T}}} \end{array}$

And,

$\begin{array}{l} \sum_{n = 0}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}} \\ \Rightarrow \sum_{n = 0}^{\infty} n {[e^{\frac{- h ω_{0}}{k_{B} T}}]}^{n} = \frac{e^{\frac{- h ω_{0}}{k_{B} T}}}{{(1 - e^{\frac{- h ω_{0}}{k_{B} T}})}^{2}} \end{array}$

Final Average Energy.

$\begin{matrix} \bar{E} = h ω_{0} \frac{e^{\frac{- h ω_{0}}{k_{B} T}}}{{(1 - e^{\frac{- h ω_{0}}{k_{B} T}})}^{2}} \frac{1}{1 - e^{\frac{- h ω_{0}}{k_{B} T}}} \\ = h ω_{0} \frac{e^{\frac{- h ω_{0}}{k_{B} T}}}{1 - e^{\frac{- h ω_{0}}{k_{B} T}}} \\ = \frac{h ω_{0}}{(e^{\frac{h ω_{0}}{k_{B} T}}) (1 - e^{\frac{- h ω_{0}}{k_{B} T}})} \\ = \frac{h ω_{0}}{e^{\frac{h ω_{0}}{k_{B} T}} - (e^{\frac{h ω_{0}}{k_{B} T}}) (e^{\frac{- h ω_{0}}{k_{B} T}})} \\ = \frac{h ω_{0}}{e^{\frac{h ω_{0}}{k_{B} T}} - 1} \end{matrix}$

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Obtain equation (9- 15) from (9-14). Make use or the following sums, correct when $| x | < 1$ :
$\begin{array}{l} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \\ \sum_{n = 0}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}} \end{array}$

Short Answer

Step by step solution

Average Energy.

Properties

Final Average Energy.

One App. One Place for Learning.

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Obtain equation (9- 15) from (9-14). Make use or the following sums, correct when |x|<1 :∑n=0∞xn=11-x∑n=0∞nxn=x(1-x)2

Short Answer

Step by step solution

Average Energy.

Properties

Final Average Energy.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Space Physics

Mechanics and Materials

Measurements

Magnetism and Electromagnetic Induction

Electricity

Kinematics Physics

Study anywhere. Anytime. Across all devices.

Obtain equation (9- 15) from (9-14). Make use or the following sums, correct when $| x | < 1$ :
$\begin{array}{l} \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \\ \sum_{n = 0}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}} \end{array}$