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Found in: Page 1272

### Fundamentals Of Physics

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

# Show that Eq. 41-9 can be written as ${{\mathbit{E}}}_{{\mathbf{F}}}{\mathbf{=}}{\mathbit{A}}{{\mathbit{n}}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$where the constant A has the value role="math" localid="1661507403881" ${\mathbf{3}}{\mathbf{.}}{\mathbf{65}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{19}}{\mathbf{}}{{\mathbf{m}}}^{{\mathbf{2}}}{\mathbf{eV}}$.

It is shown that the Energy of the Fermi level is ${E}_{\mathrm{F}}=A{n}^{2/3}$ , where the value of A is $3.65×{10}^{-19}{\mathrm{m}}^{2}\mathrm{eV}$ .

See the step by step solution

## Step 1: The given data

The equation of Fermi energy according to Eq. 41-9 is ${E}_{F}={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{{h}^{2}}{m}{n}^{2/3}$ .

where n is the number density of free electrons, m is the electron’s mass and h is Planck’s constant.

## Step 2: Understanding the concept of Fermi energy

The highest energy level occupied by the electrons in the valence band at a temperature equal to zero Kelvin is known as the Fermi level and the energy of electrons in that level is known as Fermi energy.

## Step 3: Calculation of the Fermi energy and the constant value A

From the given equation of Fermi energy, we can see that the energy is directly related to the number of conduction electrons having all other terms as constants. Thus, the formula for Femi energy can also be written as-

${E}_{F}=A{n}^{2/3}$

Where

$A={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{{\mathrm{h}}^{2}}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}={\left[\frac{3}{16\sqrt{2}\mathrm{\pi }}\right]}^{2/3}\frac{{\left(6.63×{10}^{-34}\mathrm{J}·\mathrm{s}\right)}^{2}}{9.1×{10}^{-31}\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=5.842×{10}^{-38}{\mathrm{J}}^{2}.{\mathrm{s}}^{2}/\mathrm{kg}$

Since $1\mathrm{J}=1\mathrm{kg}·{\mathrm{m}}^{2}/{\mathrm{s}}^{2}$ the units of A can also be written as-

$A=5.84×{10}^{-38}{\mathrm{m}}^{2}.\mathrm{J}$

If we divide the term by $1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}$, we have

$A=\frac{5.84×{10}^{-38}{\mathrm{m}}^{2}.\mathrm{J}}{1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}}\phantom{\rule{0ex}{0ex}}=3.65×{10}^{-19}{\mathrm{m}}^{2}\mathrm{eV}$

Hence, the value of the Fermi energy is ${E}_{F}=A{n}^{2/3}$, where the value of A is $3.65×{10}^{-19}{\mathrm{m}}^{2}\mathrm{eV}$.