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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Silver melts at ${\mathbf{961}}{\mathbf{°}}{\mathbf{C}}$. At the melting point, what fraction of the conduction electrons is in states with energies greater than the Fermi energy of 5.5 eV? (See Problem 21)

The fraction of the conduction electrons with greater energies than the Fermi energy of silver is 0.03.

See the step by step solution

## Step 1: The given data

a) Melting point of silver, $MP\left(T\right)=961°C$

b) Fermi energy of silver, ${E}_{F}=5.5 eV$

## Step 2: Understanding the concept of fraction of conduction electrons

The electrons that are free to move within the walls of the substance and are not bound to any atom, are called conduction electrons or free electrons.

Formula:

The fraction of electrons above Fermi level at the melting point are-

$frac=\frac{3\mathrm{kT}}{2{\mathrm{E}}_{\mathrm{F}}}\mathrm{where}\mathrm{k}=8.62×{10}^{-5} \mathrm{eV}/\mathrm{K}$ (i)

## Step 3: Calculation of the fraction of conduction electrons

Using the given data in equation (i), we can get the value of the fraction of the conduction electrons of the copper material as follows:

$frac=\frac{3\left(8.62×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(961+273\right)\mathrm{K}}{2\left(5.5\mathrm{eV}\right)}\phantom{\rule{0ex}{0ex}}=0.03$

Hence, the value of the fraction is 0.03.