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### Fundamentals Of Physics

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

# Show that Eq. 41-5 can be written as ${\mathbit{N}}\left(E\right){\mathbf{=}}{\mathbit{C}}{{\mathbit{E}}}^{\mathbf{1}\mathbf{/}\mathbf{2}}$. (b) Evaluate in terms of meters and electron-volts. (c) Calculate ${\mathbit{N}}\left(E\right){\mathbf{}}{\mathbf{for}}{\mathbf{}}{\mathbit{E}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{eV}}$ for .

1. The density of the states of conduction electrons is $N\left(E\right)=C{E}^{1/2}$.
2. The value of the constant in meters and electron-volts is $6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}$.
3. The density of states for is $E=5\mathrm{eV}\mathrm{is}1.52×{10}^{28}{\mathrm{eV}}^{-1}.{\mathrm{m}}^{-3}$.
See the step by step solution

## Step 1: The given data

1. The given energy value, $E=5eV$

## Step 2: Understanding the concept of density of states

Using the formula of the density of states associated with the conduction electrons of metal, we can get the required value of the density of the state for the given energy and constant values.

${\mathbf{N}}\left(E\right){\mathbf{=}}\frac{\mathbf{8}\sqrt{\mathbf{2}}{\mathbf{\tau \tau m}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{{\mathbf{h}}^{\mathbf{3}}}{{\mathbf{E}}}^{\mathbf{1}\mathbf{/}\mathbf{2}}$

## Step 3: a) Calculation of the formula of density of states

From the given equation of density of states of the conduction electrons of a metal, we get that $N\left(E\right)=C{E}^{1/2}....................\left(a\right)$

where

$C=\frac{8\sqrt{2}{\mathrm{\pi m}}^{3/2}}{{h}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{8\sqrt{2}\mathrm{\pi }{\left(9.1×{10}^{-31}\mathrm{kg}\right)}^{3/2}}{{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}^{3}}\phantom{\rule{0ex}{0ex}}=1062×{10}^{56}{\mathrm{kg}}^{3/2}{\mathrm{J}}^{3}.{\mathrm{s}}^{3}$

Hence, the density of the states of conduction electrons is $N\left(E\right)=C{E}^{1/2}$.

## Step 4: b) Calculation of the value of constant C in meters and electron-volts

We know that $1\mathrm{J}=1\mathrm{kg}.{\mathrm{m}}^{2}/{\mathrm{s}}^{2}$

So, considering the equation of kinetic energy $\left(K=\frac{1}{2}m{v}^{2}\right)$ the unit of mass is $1\mathrm{kg}=1\mathrm{J}.{\mathrm{s}}^{2}{\mathrm{m}}^{-2}$.

Thus, the units of C becomes-

${\left(\mathrm{J}.{\mathrm{s}}^{2}\right)}^{3/2}.{\left({\mathrm{m}}^{-2}\right)}^{3/2}.{\mathrm{J}}^{-3}.{\mathrm{s}}^{-3}={\mathrm{J}}^{-3/2}{\mathrm{m}}^{-3}$

Now, the value of C in meters and electron-volts can be given as:

$\mathrm{C}=\left(1.062×{10}^{56}{\mathrm{kg}}^{3/2}/{\mathrm{J}}^{3}.{\mathrm{s}}^{3}\right){\left(1.602×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}^{3/2}\phantom{\rule{0ex}{0ex}}=6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}$

Hence, the value of C is $6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}$.

## Step 5: c) Calculation of the density of states for the given energy

Using the given energy value in equation (a) of density in part (a), we can get the density of the states of the conduction electrons of the metal as follows:

$\mathrm{N}\left(\mathrm{E}\right)=\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}{\left(5\mathrm{eV}\right)}^{1/2}\right)\phantom{\rule{0ex}{0ex}}=1.52×{10}^{28}{\mathrm{eV}}^{-1}.{\mathrm{m}}^{-3}$

Hence, the value of the density of electrons is $1.52×{10}^{28}{\mathrm{eV}}^{-1}.{\mathrm{m}}^{-3}$.