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Found in: Page 1273

### Fundamentals Of Physics

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

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# The Fermi energy of aluminum is 11.6 eV; its density and molar mass are${\mathbf{2}}{\mathbf{.}}{\mathbf{70}}{\mathbf{}}{\mathbf{g}}{\mathbf{/}}{{\mathbf{cm}}}^{{\mathbf{3}}}$ and ${\mathbf{2}}{\mathbf{.}}{\mathbf{70}}{\mathbf{}}{\mathbf{g}}{\mathbf{/}}{\mathbf{mol}}$, respectively. From these data, determine the number of conduction electrons per atom.

The number of conduction electrons contributed by an atom of aluminum is 3.

See the step by step solution

## Step 1: The given data

a) Fermi energy of aluminum,${E}_{F}=11.6 eV$

b) Density of aluminum,$d=2.7 \mathrm{g}/{\mathrm{cm}}^{3}$

c) Molar mass of aluminum,$A=27 \mathrm{g}/\mathrm{mol}$

## Step 2: Understanding the concept of conduction electrons per atom

The electrons that jump from the valence band to the conduction band, by absorbing energy from the surrounding, are called conduction electrons. These electrons are now free to move within the walls of the sample of a substance.

Formulae:

The Energy of Fermi level of a metal,${E}_{F}=A{n}^{2/3} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

where, is the number density of conduction electrons and $A=3.65×{10}^{-19} {m}^{2}eV$

The number of atoms per unit volume,$N=d/M \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Where, d = density and M is the mass per atom$M=A/{N}_{A}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

The mass of a substance per atom,

$where{N}_{A}=6.022×{10}^{23} /mol$ and A is the molar mass.

## Step 3: Calculation of the number of conduction electrons per atom

Let N be the number of atoms per unit volume and n be the number of conduction electrons per unit volume.

At first, the number of conduction electrons per unit volume of aluminum can be calculated using equation as follows:

$n={\left(\frac{{E}_{F}}{A}\right)}^{3/2}\phantom{\rule{0ex}{0ex}}={\left(\frac{11.6eV}{3.65×{10}^{-19}{m}^{2}eV}\right)}^{3/2}\phantom{\rule{0ex}{0ex}}=1.79×{10}^{29}{m}^{-3}........................\left(a\right)$

Now, the mass of aluminum per atom can be calculated using equation and the given data, as follows:

$M=\frac{27g/mol}{6.022×{10}^{23}/mol}\phantom{\rule{0ex}{0ex}}=4.48×{10}^{-23}g$

Now, using this mass value in equation , we can get the value of the number of atoms per unit volume as follows:

$N=\frac{2.7g/c{m}^{3}}{4.48×{0}^{-23}g}\phantom{\rule{0ex}{0ex}}=6.03×{10}^{22}/c{m}^{3}\phantom{\rule{0ex}{0ex}}=6.03×{10}^{28}/{m}^{3}.............................\left(b\right)$

Now, the number of conduction electrons per atom can be calculated by dividing equation (a) by equation (b) as follows:

$\frac{n}{N}=\frac{1.79×{10}^{29}{m}^{-3}}{6.03×{10}^{28}/{m}^{3}}\phantom{\rule{0ex}{0ex}}=2.97\phantom{\rule{0ex}{0ex}}\approx 3$

Hence, the required number of conduction electrons per atom is 3.

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