Suggested languages for you:

Americas

Europe

Q24P

Expert-verified
Found in: Page 1273

Fundamentals Of Physics

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

A certain material has a molar mass of 20.0g/mol , Fermi energy of 5.00 eV , and 2 valence electrons per atom. What is the density ${\mathbf{\left(}}{\mathbf{g}}{\mathbf{/}}{{\mathbf{cm}}}^{{\mathbf{3}}}{\mathbf{\right)}}$?

The density of the material is $0.84\mathrm{g}/{\mathrm{cm}}^{3}$ .

See the step by step solution

Step 1: The given data

1. Molar mass of the material,A = 20 g/mol .
2. Fermi energy of the material, ${\mathrm{E}}_{\mathrm{F}}=5\mathrm{eV}$.
3. Each atom has 2 valence electrons.

Step 2: Understanding the concept of density

The number of conduction electrons per unit volume per unit energy range, at a particular energy is given as number density of these conduction electrons.

The equation of Fermi energy is

${{\mathbf{E}}}_{{\mathbf{F}}}{\mathbf{=}}{\left[\frac{3}{16\sqrt{2}\mathrm{\tau \tau }}\right]}^{\mathbf{2}\mathbf{/}\mathbf{3}}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{m}}{{\mathbf{n}}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ (i)

where, n is the number of conduction electrons per unit volume, m is the mass of an electron and h is the Planck’s constant.

The density of a material according to the number of atoms per unit volume and molar mass of a material,

${{\mathbf{P}}}_{{\mathbf{material}}}{\mathbf{=}}{{\mathbf{n}}}_{{\mathbf{atoms}}}{\mathbf{A}}$ (ii)

Step 3: Calculation of the density of the material

At first using equation (i) and the given value of Fermi energy, we calculate the number of conduction electrons per unit volume as follows:

$\mathrm{n}=\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{E}}_{\mathrm{F}}}{{\mathrm{h}}^{2}}\right)}^{3/2}\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^{2}{\mathrm{E}}_{\mathrm{F}}}{{\left(\mathrm{hc}\right)}^{2}}\right)}^{3/2}\phantom{\rule{0ex}{0ex}}=\frac{16\sqrt{2}\mathrm{\pi }}{3}{\left(\frac{\left(0.511×{10}^{6}\mathrm{eV}\right)\left(5\mathrm{eV}\right)}{{\left(1240\mathrm{eV}.\mathrm{nm}\right)}^{2}}\right)}^{3/2}\left(\because {\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^{2}=0.511×{10}^{6}\mathrm{eV},\mathrm{hc}=1240\mathrm{eV}.\mathrm{nm}\right)\phantom{\rule{0ex}{0ex}}=50.9/{\mathrm{nm}}^{3}$

The number of moles per unit volume is given as-

$\mathrm{n}=\frac{5.09×{10}^{28}/{\mathrm{m}}^{3}}{6.022×{10}^{23}/\mathrm{mol}}\left(\mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number}=6.022×{10}^{23}/\mathrm{mol}\right)\phantom{\rule{0ex}{0ex}}=8.4×{10}^{4}\mathrm{mol}/{\mathrm{m}}^{3}$

Now, we are given the atoms are bivalent,

${\mathrm{n}}_{\mathrm{atom}}=\mathrm{n}/2\phantom{\rule{0ex}{0ex}}=\left(8.4×{10}^{4}\mathrm{mol}/{\mathrm{m}}^{3}\right)/2\phantom{\rule{0ex}{0ex}}=4.2×{10}^{4}\mathrm{mol}/{\mathrm{m}}^{3}$

Thus, using this value in equation (ii), we can get the density of the material as follows:

${\mathrm{P}}_{\mathrm{material}}=\left(4.2×{10}^{4}\mathrm{mol}/{\mathrm{m}}^{3}\right)\left(20\mathrm{g}/\mathrm{mol}\right)\phantom{\rule{0ex}{0ex}}=8.4×{10}^{5}\mathrm{g}/{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}=0.84\mathrm{g}/{\mathrm{cm}}^{3}$

Hence, the value of the density is 0.84 $\mathrm{g}/{\mathrm{cm}}^{3}$ .