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Q5P

Expert-verifiedFound in: Page 1273

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

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

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

- The given energy value, $E=5eV$

**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.**

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

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

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

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

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:

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

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

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:

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

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

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