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Q1P

Expert-verifiedFound in: Page 1272

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

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

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

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.

**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.**

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

$A={\left[\frac{3}{16\sqrt{2}\mathrm{\pi}}\right]}^{2/3}\frac{{\mathrm{h}}^{2}}{\mathrm{m}}\phantom{\rule{0ex}{0ex}}={\left[\frac{3}{16\sqrt{2}\mathrm{\pi}}\right]}^{2/3}\frac{{\left(6.63\times {10}^{-34}\mathrm{J}\xb7\mathrm{s}\right)}^{2}}{9.1\times {10}^{-31}\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=5.842\times {10}^{-38}{\mathrm{J}}^{2}.{\mathrm{s}}^{2}/\mathrm{kg}$

Since $1\mathrm{J}=1\mathrm{kg}\xb7{\mathrm{m}}^{2}/{\mathrm{s}}^{2}$ the units of A can also be written as-

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

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

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

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

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