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### Fundamentals Of Physics

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

# Silver is a monovalent metal. Calculate (a) the number density of conduction electrons, (b) the Fermi energy, (c) the Fermi speed and (d) the de Broglie wavelength corresponding to this electron speed. See Appendix F for the needed data on silver.

a) The number density of conduction electrons in silver is $5.86×{10}^{28}{\mathrm{m}}^{-3}$ .

b) The Fermi energy of silver metal is $5.49\mathrm{eV}$ .

c) The Fermi speed of silver metal is $1.39×{10}^{6}\mathrm{m}/\mathrm{s}$ .

d) The de-Broglie wavelength corresponding to this electron speed is $5.22×{10}^{-10}\mathrm{m}$ .

See the step by step solution

## Step 1: The given data

a) The monovalent element silver is given.

b) Molar mass of silver (Appendix F), A = 107.870 g/mol

c) Density of silver (Appendix F), d = 10.49 g/${\mathrm{cm}}^{3}$

## Step 2: Understanding the concept of density and molar mass

Using the given formula for the number of atoms per unit volume, we can get the required value of the conduction electrons per unit volume considering the metal is monovalent. Now using this value in the equation of Fermi energy, we calculated the energy of the metal. Now, using this value of Fermi energy, we can get the speed of the electron in conduction. Further, we can use this value of speed in the de-Broglie wavelength relation; we can get the value of the wavelength of the electron.

Formulae:

• The mass of an atom, $M=A/{N}_{A},\mathrm{where}{N}_{A}=6.022×{10}^{23}{\mathrm{mol}}^{-1}$ (i)
• The number density of conduction electrons, role="math" localid="1661926845725" $n=\frac{d}{M}$ (ii)

d= density of the atom, M = mass of a single atom

• The equation of Fermi energy ${E}_{F}=\frac{0.121{h}^{2}}{m}{n}^{2/3}$ (iii)

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

• The energy equation of a moving body, $E=\frac{1}{2}m{v}^{2}$ (iv)
• The de-Broglie wavelength according to an electron speed,${\lambda }_{F}=\frac{h}{m{v}_{F}}$ (v)

Here, ${v}_{F}$ is the fermi speed.

## Step 3: a) Calculation of number density of conduction electrons in silver

Since each atom contributes one conduction electron, using equation (i) and equation (ii), we can get the number density of the conduction electrons in silver as follows:

$n=\frac{d}{A/{N}_{A}}\phantom{\rule{0ex}{0ex}}=\frac{\left(10.49\mathrm{g}/{\mathrm{cm}}^{2}\right)\left({10}^{6}{\mathrm{cm}}^{3}/{\mathrm{m}}^{3}\right)}{\left(107.870\mathrm{g}/\mathrm{mol}\right)/\left(6.022×{10}^{23}{\mathrm{mol}}^{-1}\right)}\phantom{\rule{0ex}{0ex}}=5.86×{10}^{28}{\mathrm{m}}^{-3}$

Hence, the value of the number density is $5.86×{10}^{28}{\mathrm{m}}^{-3}$.

## Step 4: b) Calculation of the Fermi energy

Using the given data and the above number density value in equation (iii), the Fermi energy of silver metal can be calculated as follows:

${E}_{F}=\frac{0.121{\left(6.63×{10}^{-34}\mathrm{J},\mathrm{s}\right)}^{2}}{9.1×{10}^{-31}\mathrm{kg}}{\left(5.86×{10}^{28}{\mathrm{m}}^{-3}\right)}^{2/3}\phantom{\rule{0ex}{0ex}}=8.80×{10}^{-19}\mathrm{J}\phantom{\rule{0ex}{0ex}}=5.49\mathrm{eV}$

Hence, the value of Fermi energy is 5.49 eV.

## Step 5: c) Calculation of the Fermi speed

Using the above Fermi energy value and equation (iv), we can get the speed of the conducting electron in Fermi level of silver as follows:

${V}_{F}=\sqrt{\frac{2{E}_{F}}{m}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2\left(8.80×{10}^{-19}\mathrm{J}\right)}{\left(9.1×{10}^{-31}\mathrm{kg}\right)}}\phantom{\rule{0ex}{0ex}}=1.39×{10}^{6}\mathrm{m}/\mathrm{s}$

Hence, the value of Fermi speed is $1.39×{10}^{6}\mathrm{m}/\mathrm{s}$.

## Step 6: d) Calculation of the de-Broglie wavelength

Using the above speed value, the de-Broglie wavelength corresponding to the speed of the conducting electron can be given using equation (v) as follows:

${\lambda }_{F}=\frac{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}{\left(9.1×{10}^{-31}\mathrm{kg}\right)\left(1.39×{10}^{6}\mathrm{m}/\mathrm{s}\right)}\phantom{\rule{0ex}{0ex}}=5.22×{10}^{-10}\mathrm{m}$

Hence, the value of de-Broglie wavelength is $5.22×{10}^{-10}\mathrm{m}$.