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Found in: Page 1272

### Fundamentals Of Physics

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

# Calculate the density of states ${\mathbit{N}}{\mathbf{\left(}}{\mathbit{E}}{\mathbf{\right)}}$ for metal at energy ${\mathbit{E}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{0}}{\mathbit{e}}{\mathbit{V}}$and show that your result is consistent with the curve of Fig. 41-6.

The density of states $N\left(E\right)$ for metal is $1.9×{10}^{28}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-1}$ and it is consistent with the curve of figure 41-6.

See the step by step solution

## Step 1: The given data

Energy of the metal, $E=8\mathrm{eV}$

## Step 2: the concept of density of states

The number of states per unit energy range per unit volume $\left(N\left(E\right)\right)$, present in a sample of the material at a particular energy $\left(E\right)$, is known as density of states. The formula for density of states is given as-

$N\left(E\right)=\frac{8\sqrt{2}{\mathrm{\pi m}}^{3/2}}{{h}^{3}}{E}^{1/2}.............................\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{where}\mathrm{h}=6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{and}\mathrm{m}=9.1×{10}^{-31}\mathrm{kg}$

## Step 3: Calculation of the density of states of a metal

We can write equation (1) as follows:

$N\left(E\right)=C{E}^{1/2}$

In the above equation, the value of C is -

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

Using the given data in equation (1), the density of states for the metal with energy $E=8eV$can be calculated as follows:

$N\left(E\right)=\left[6.81×{10}^{27}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-2/3}\right]{\left(8\mathrm{eV}\right)}^{1/2}\phantom{\rule{0ex}{0ex}}=1.9×{10}^{28}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-1}$

This value of density of state is consistent with the given figure 41-6.

Hence, the value of the density of states is $1.9×{10}^{28}{\mathrm{m}}^{-3}.{\left(\mathrm{eV}\right)}^{-1}$.