What is the Density of Conduction Electrons in Monovalent Copper?

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SUMMARY

The density of conduction electrons in monovalent copper is calculated to be approximately 8.4375 x 1028 m-3. This value is derived using the formula n = (D * NA) / m, where D is the density (9000 kg m-3), NA is Avogadro's number (6.02 x 1023 mol-1), and m is the atomic mass (64 amu). The calculation confirms that the density of conduction electrons is directly related to the material's density and atomic mass, providing a clear understanding of electron distribution in copper.

PREREQUISITES
  • Understanding of Fermi-Dirac statistics
  • Knowledge of atomic mass units (amu)
  • Familiarity with Avogadro's number
  • Basic principles of density calculations
NEXT STEPS
  • Study the derivation and applications of Fermi-Dirac statistics
  • Learn about the concept of Fermi energy in metals
  • Explore the relationship between electron density and material properties
  • Investigate the implications of conduction electron density on electrical conductivity
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Students and researchers in physics, materials science, and electrical engineering who are studying the properties of conductive materials, particularly metals like copper.

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Homework Statement



Monovalent copper (one conduction electron per atom) has a density of 9000 kg m-3 and atomic mass of 64 amu (ie. 1 kmole = 64kg). Find:

a)The density of conduction electrons per unit volume

b)the Fermi energy in electron volts

Homework Equations



f(E) = 1/(e(E-EF)/KT+1)

ρ=m/V

The Attempt at a Solution



So this is the first problem I've done on Fermi-Dirac so I'm a bit stuck here.

So I got the volume from V=m/ρ, so:

V=64/9000=.007m3

Is this even relevant to the question? Really lost and any help would be much appreciated
 
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I think I have to use the formula n=DNA/m

where:

NA=Avogadros No. = 6.02x1023 mol-1
D=Density=9000 kg m-3=9000000 g m-3
m=64 amu= g mol-1

I used the formula to get n=(9000000)(6.02x1023)/64=8.4375x1028 m-3

Is this the answer for part a)?
 

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