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Answers without the blur. Sign up and see all textbooks for free! Q. 2.15

Expert-verified Found in: Page 63 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Use a pocket calculator to check the accuracy of Stirling's approximation for $N=50$ . Also check the accuracy of equation $2.16$ for $\mathrm{ln}N!$ .

By using pocket calculator as,

$50!=3.0414×{10}^{64};\mathrm{ln}\left(50!\right)=148.4778$ and

By using Stirling's approximation as,

$50!\approx 3.0363×{10}^{64,};\mathrm{ln}\left(50!\right)\approx 145.6012.$

See the step by step solution

## Step: 1 Using pocket calculator:

By using pocket calculator $n=50$ as

$\begin{array}{l}N!=50!=3.0414×{10}^{64}\\ \mathrm{ln}\left(N!\right)=\mathrm{ln}\left(50!\right)=148.4778\end{array}$

We have,

$\begin{array}{l}N!=50!\approx {50}^{\left(}50\right){e}^{-50}\sqrt{2\pi 50}\\ N!=3.0363×{10}^{64}\\ 50!\approx 3.0363×{10}^{64}.\end{array}$

## Step: 2 Using Stirling's approximation:

By using Stirling's approximation as,

$\begin{array}{l}N!\approx {N}^{N}{e}^{-N}\sqrt{2\pi N}\\ \mathrm{ln}\left(N!\right)\approx Nln\left(N\right)-N\end{array}$

We have,

$\begin{array}{l}\mathrm{ln}\left(N!\right)=\mathrm{ln}\left(50!\right)\approx 50\mathrm{ln}\left(50\right)-50\\ ln\left(N!\right)=145.6012\\ \mathrm{ln}\left(50!\right)\approx 145.6012\end{array}$

The Stirling's approximation is applicable for even $n=50.$

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