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Expert-verified Found in: Page 1273 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Calculate ${{\mathbit{N}}}_{{\mathbf{0}}}\left(E\right)$ the density of occupied states, for copper at 10000K for energy E of (a)4.00eV , (b) 6.75eV, (c) 7.00eV, (d) 7.25eV, and (e) 9.00eV. Compare your results with the graph of Fig. 41-8b. The Fermi energy for copper is 7.00eV. a) The density of occupied states for energy $4\mathrm{eV}\mathrm{is}1.36×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$ .

b) The density of occupied states for energy $6.75\mathrm{eV}\mathrm{is}1.68×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$ .

c) The density of occupied states for energy $7\mathrm{eV}\mathrm{is}9.01×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$ .

d) The density of occupied states for energy $7.25\mathrm{eV}\mathrm{is}9.56×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$ .

e) The density of occupied states for energy $9\mathrm{eV}\mathrm{is}1.71×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

See the step by step solution

## Step 1: The given data

a) The Fermi energy for copper, ${E}_{F}=7\mathrm{eV}$

b) The temperature of the energy state of copper, E = 1000 K

c) The energy states are 4 eV , 6.75 eV , 7 eV ,7.25 eV and 9 eV .

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

Using the concept of probability occupancy and the density of states, the required value of the densities of the conduction electrons for copper at different energy levels can be calculated substituting the given data.

Formulae:

The density of states associated with the conductions electrons of a metal ${\mathbf{N}}\left(E\right){\mathbf{=}}{{\mathbf{CE}}}^{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{}}{\mathbf{,}}{\mathbf{}}{\mathbf{where}}{\mathbf{}}{\mathbf{C}}{\mathbf{=}}\frac{\mathbf{8}\sqrt{\mathbf{2}}{\mathbf{\tau \tau m}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}{{\mathbf{h}}^{\mathbf{3}}}$(i)

The probability of the condition that a particle will have energy E, (ii) ${\mathbf{P}}\left(E\right){\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{e}}^{\left(E-{E}_{F}\right)\mathbf{/}\mathbf{kT}}\mathbf{+}\mathbf{1}}{\mathbf{,}}{\mathbf{where}}{\mathbf{}}{\mathbf{k}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{62}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{5}}{\mathbf{eV}}$

The density of states for the condition of occupancy by the conduction electrons,

${{\mathbf{N}}}_{{\mathbf{0}}}\left(E\right){\mathbf{=}}{\mathbf{N}}\left(E\right){\mathbf{P}}\left(E\right)$ (iii)

## Step 3: a) Calculation of the density of occupied states for 4 eV

At first, the density of the occupied states can be given using equations (i) and (ii) in equation (iii) can be given as follows:

${N}_{0}\left(E\right)=C{E}^{1/2}{\left[{e}^{\left(E-{E}_{F}\right)/kT}+1\right]}^{-1}......................\left(\mathrm{a}\right)$

where,

$C=\frac{8\sqrt{2}\pi {m}^{3/2}}{{h}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{8\sqrt{2}\pi {\left(9.1×{10}^{-31}\mathrm{kg}\right)}^{3/2}}{{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}^{3}}\phantom{\rule{0ex}{0ex}}=1.062×{10}^{56}{\mathrm{kg}}^{3/2}{\mathrm{J}}^{3}.{\mathrm{s}}^{3}\phantom{\rule{0ex}{0ex}}=6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}$

Thus, the density of occupied states at E = 4eV is given using equation (a) as follows:

role="math" localid="1661929491044" ${\mathrm{N}}_{0}\left(\mathrm{E}\right)=\frac{\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right){\left(4\mathrm{eV}\right)}^{1/2}}{\left[\mathrm{exp}\left(\left(4\mathrm{eV}-7\mathrm{eV}\right)/\left(8.62×{10}^{-5}\mathrm{eV}\right)\left(1000\mathrm{K}\right)\right)+1\right]}\phantom{\rule{0ex}{0ex}}=1.36×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$

Hence, the value of the density of states is $1.36×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

## Step 4: b) Calculation of the density of occupied states for 6.75 eV

Thus, the density of occupied states at E = 6.75 eV is given using equation (a) as follows:

${\mathrm{N}}_{0}\left(\mathrm{E}\right)=\frac{\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right){\left(6.75\mathrm{eV}\right)}^{1/2}}{\left[\mathrm{exp}\left(\left(6.75\mathrm{eV}-7\mathrm{eV}\right)/\left(8.62×{10}^{-5}\mathrm{eV}\right)\left(1000\mathrm{K}\right)\right)+1\right]}\phantom{\rule{0ex}{0ex}}=1.68×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$

Hence, the value of the density of states is $1.68×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

## Step 5: c) Calculation of the density of occupied states for 7 eV

Thus, the density of occupied states at E = 7 eV is given using equation (a) as follows:

${\mathrm{N}}_{0}\left(\mathrm{E}\right)=\frac{\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right){\left(7\mathrm{eV}\right)}^{1/2}}{\left[\mathrm{exp}\left(\left(7\mathrm{eV}-7\mathrm{eV}\right)/\left(8.62×{10}^{-5}\mathrm{eV}\right)\left(1000\mathrm{K}\right)\right)+1\right]}\phantom{\rule{0ex}{0ex}}=9.01×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$

Hence, the value of the density of states is $9.01×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

## Step 6: d) Calculation of the density of occupied states for 7.25 eV

Thus, the density of occupied states at E = 7.25 eV is given using equation (a) as follows:

${\mathrm{N}}_{0}\left(\mathrm{E}\right)=\frac{\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right){\left(7.25\mathrm{eV}\right)}^{1/2}}{\left[\mathrm{exp}\left(\left(7.25\mathrm{eV}-7\mathrm{eV}\right)/\left(8.62×{10}^{-5}\mathrm{eV}\right)\left(1000\mathrm{K}\right)\right)+1\right]}\phantom{\rule{0ex}{0ex}}=9.56×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$

Hence, the value of the density of states is $9.56×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$.

## Step 7: e) Calculation of the density of occupied states for 9 eV

Thus, the density of occupied states at E = 9 eV is given using equation (a) as follows:

${\mathrm{N}}_{0}\left(\mathrm{E}\right)=\frac{\left(6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right){\left(9\mathrm{eV}\right)}^{1/2}}{\left[\mathrm{exp}\left(\left(9\mathrm{eV}-7\mathrm{eV}\right)/\left(8.62×{10}^{-5}\mathrm{eV}\right)\left(1000\mathrm{K}\right)\right)+1\right]}\phantom{\rule{0ex}{0ex}}=1.71×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$

Hence, the value of the density of states is $1.71×{10}^{28}{\mathrm{m}}^{-3}.{\mathrm{eV}}^{-1}$ .

All the values are compared to the graph and consistent with the graph values. ### Want to see more solutions like these? 