Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Assume that the total volume of a metal sample is the sum of the volume occupied by the metal ions making up the lattice and the (separate) volume occupied by the conduction electrons. The density and molar mass of sodium (a metal) are $971 kg / m^{3}$ and $23 . g / mol$ , respectively; assume the radius of the Na+ ion is . (a) What percent of the volume of a sample of metallic sodium is occupied by its conduction electrons? (b) Carry out the same calculation for copper, which has density, molar mass, and ionic radius of 8960 $kg / m^{3}$ , 63.5g/mol, and 135 pm, respectively. (c) For which of these metals do you think the conduction electrons behave more like a free-electron gas?

The percentage of the volume, the conduction electrons of a sample of metallic sodium occupy is 90 %.
The percentage of the volume, the conduction electrons of a sample of metallic copper occupy is 12.4 %.

The metal whose conduction electrons behave more like a free-electron gas is sodium

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Density of sodium, $d_{sodium} = 971 \lg / m^{3}$
Molar mass of sodium, $A_{sodium} = 23.0 g / mol$
Radius of sodium ion, $R_{sodium} = 98 pm$
Density, molar mass and ionic radius of copper are:

$d_{copper} = 8960 kg / m^{3}, A_{copper} = 63.5 g / mol, R_{copper} = 135 pm$

Step 2: Understanding the concept of volume occupied by the conduction electrons

Using the concept of molar mass, we can get the volume per cubic meter occupied by the conduction electrons in a metal. Now, using this value in the fractional formula, we can get the required ratio or percentage of the conduction electrons.

Formulae:

The volume per cubic meter of a substance,

$V (volume / {cm}^{3}) = \frac{{dN}_{A}}{A} (\frac{4}{3} {πR}^{3}), where N_{A} = 6.022 \times 10^{23} / mol$ (i)

The fraction of volume available for conduction electrons, $Fraction = 1 - \frac{V_{substance}}{1.00 m^{3}}$ (ii)

Step 3: a) Calculation of the fraction of volume available for conduction electrons of sodium

Using the given data in equation (i), the volume per cubic meter of sodium occupied by the sodium ions can be given as follows:

$V (volume / {cm}^{3}) = \frac{(971 kg / m^{3}) (6.022 \times 10^{23} / mol)}{(23.0 \times 10^{- 3} kg / mol)} (\frac{4}{3} π {(98.0 \times 10^{- 12} m)}^{3}) = 0.100 m^{3}$

Now, the fraction of volume available for conduction electrons can be given using the above value in equation (ii) as follows:

$Fraction = 1 - \frac{0.100 m^{3}}{1.00 m^{3}} = 0.900 or 90.0 %$

Hence, the value of the occupancy percentage is 90.0 % .

Step 4: b) Calculation of the fraction of volume available for conduction electrons of copper

Using the given data in equation (i), the volume per cubic meter of sodium occupied by the sodium ions can be given as follows:

$V (volume / {cm}^{3}) = \frac{(8960 kg / m^{3}) (6.022 \times 10^{23} / mol)}{(63.5 \times 10^{- 3} kg / mol)} (\frac{4}{3} π {(135.0 \times 10^{- 12} m)}^{3}) = 0.1876 m^{3}$

Now, the fraction of volume available for conduction electrons can be given using the above value in equation (ii) as follows:

$Fraction = 1 - \frac{0.1876 m^{3}}{1.00 m^{3}} = 0.124 or 12.4 %$

Hence, the value of the occupancy percentage is 12.4 % .

Step 5: c) Calculation of the metal whose conduction electrons behave more like free-electron gas

From the above calculations, the metal whose conduction electrons are more like free-electron gas is sodium because the electrons occupy a greater portion of the space available.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of volume occupied by the conduction electrons

Step 3: a) Calculation of the fraction of volume available for conduction electrons of sodium

Step 4: b) Calculation of the fraction of volume available for conduction electrons of copper

Step 5: c) Calculation of the metal whose conduction electrons behave more like free-electron gas

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of volume occupied by the conduction electrons

Step 3: a) Calculation of the fraction of volume available for conduction electrons of sodium

Step 4: b) Calculation of the fraction of volume available for conduction electrons of copper

Step 5: c) Calculation of the metal whose conduction electrons behave more like free-electron gas

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