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.

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fluids

Circular Motion and Gravitation

Quantum Physics

Kinematics Physics

Magnetism

Electricity

Study anywhere. Anytime. Across all devices.

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

Fluids

Circular Motion and Gravitation

Quantum Physics

Kinematics Physics

Magnetism

Electricity

Study anywhere. Anytime. Across all devices.

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