Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $(g / {cm}^{3})$ ?

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Molar mass of the material,A = 20 g/mol .
Fermi energy of the material, $E_{F} = 5 eV$ .
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

$E_{F} = {[\frac{3}{16 \sqrt{2} ττ}]}^{2 / 3} \frac{h^{2}}{m} n^{2 / 3}$ (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,

$P_{material} = n_{atoms} A$ (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:

$n = \frac{16 \sqrt{2} π}{3} {(\frac{m_{e} E_{F}}{h^{2}})}^{3 / 2} = \frac{16 \sqrt{2} π}{3} {(\frac{m_{e} c^{2} E_{F}}{{(hc)}^{2}})}^{3 / 2} = \frac{16 \sqrt{2} π}{3} {(\frac{(0.511 \times 10^{6} eV) (5 eV)}{{(1240 eV . nm)}^{2}})}^{3 / 2} (∵ m_{e} c^{2} = 0.511 \times 10^{6} eV, hc = 1240 eV . nm) = 50.9 / {nm}^{3}$

The number of moles per unit volume is given as-

$n = \frac{5.09 \times 10^{28} / m^{3}}{6.022 \times 10^{23} / mol} (Avogadro' s number = 6.022 \times 10^{23} / mol) = 8.4 \times 10^{4} mol / m^{3}$

Now, we are given the atoms are bivalent,

$n_{atom} = n / 2 = (8.4 \times 10^{4} mol / m^{3}) / 2 = 4.2 \times 10^{4} mol / m^{3}$

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

$P_{material} = (4.2 \times 10^{4} mol / m^{3}) (20 g / mol) = 8.4 \times 10^{5} g / m^{3} = 0.84 g / {cm}^{3}$

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

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Fundamentals Of Physics

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Short Answer

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 $(g / {cm}^{3})$ ?

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of density

Step 3: Calculation of the density of the material

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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 (g/cm3)?

Step by Step Solution

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