Finding the number of free electrons

  • #1
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Homework Statement:
Prove that the concentration n of free electrons per cubic meter of a metal is given by
n = d*v/AM = A0dv*10^3/A
where d = density, kg/m^3
v = valence, free electrons per atom
A = atomic weight
M = weight of atom of unit atomic weight, kg
A0 = Avogadro's number, molecules/mole
Relevant Equations:
1gram = A*1amu
The required number of free electrons to be calculated is
##n = \frac{dv} {AM} = \frac{A_0dv*10^3} {A} ##
I know mass of single atom is ##M##
The atomic weight of mole of atoms is ##A##.
Hence ##A = A_0*M ##
##d = \frac{A} {1} ## -> "1" represents ##1m^3##. I am confused how to proceed further. How do i approach?
 

Answers and Replies

  • #2
PeroK
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Homework Statement:: Prove that the concentration n of free electrons per cubic meter of a metal is given by
n = d*v/AM = A0dv*10^3/A
where d = density, kg/m^3
v = valence, free electrons per atom
A = atomic weight
M = weight of atom of unit atomic weight, kg
A0 = Avogadro's number, molecules/mole
Relevant Equations:: 1gram = A*1amu

The required number of free electrons to be calculated is
##n = \frac{dv} {AM} = \frac{A_0dv*10^3} {A} ##
I know mass of single atom is ##M##
The atomic weight of mole of atoms is ##A##.
Hence ##A = A_0*M ##
##d = \frac{A} {1} ## -> "1" represents ##1m^3##. I am confused how to proceed further. How do i approach?
If you have the density (in ##kg/m^3##), then you need to calculate now many atoms are in a cubic metre. That means how many atoms there are in a ##kg##. Can you calculate that?

Or, alternatively, what is the mass of an atom in ##kg##?
 
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  • #3
PeroK
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Relevant Equations:: 1gram = A*1amu
This should be ##1g = A_0 \times 1 amu##.

That means that ##A_0## things of mass ##1 amu## have a mass of ##1g##.
 
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  • #4
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After some struggle, this is how i approached
Or, alternatively, what is the mass of an atom in ##kg##?
Mass of an atom in kg is ##\frac{A*10^{-3}} {A_0} kg## ----> eq1
If the number of atoms in ##1m^3## of volume is ##n##.
Then the mass of the total atoms is ##\frac{nA*10^{-3}} {A_0}## ----> eq2
##density = \frac{mass} {volume};## volume = ##1m^3##
##d = \frac{nA*10^{-3}} {1* A_0}## ----> eq3
Hence the number of atoms
##n =\frac{dA_0*10^3} {A} ## ---> eq4.
If there are ##v## valence electrons for each atom then the concentration of free electrons is
##n =\frac{dvA_0*10^3} {A}## ---> eq5
 
  • #5
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Okay, that's the answer, but it looks complicated and you've used ##n## as the number of atoms and the number of valence electrons. Let's use ##N## for the number of atoms per cubic metre and ##m## for the mass of a single atom.

First, we know, by definition, that $$m = A \ amu$$ And we also know that $$M \ kg = 1 \ amu$$ because you are told that an atom if ##1 \ amu## has a mass of ##M \ kg##. This gives us the mass of an atom in ##kg##: $$m = AM \ kg$$ The number of atoms in ##d \ kg## is: $$N = \frac{d}{AM}$$
Alternatively, we know by definition that:

##A_0## is the number of atoms of ##1 \ amu## in ##1g##

and ##A_0 \times 10^3## is the number of atoms of ##1 \ amu## in one ##kg##,

hence ##\frac{ A_0 \times 10^3}{A}## is the number of atoms of ##A \ amu## in one ##kg##,

and ##N = \frac{d \ A_0 \times 10^3}{A}## is the number of atoms of ##A \ amu## in ##d \ kg##
 
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  • #6
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Thank you for the help, the solution now looks more in order. but it is confusing ##1 \text{ amu}, A\text{ amu} ## etc. They could have started directly defining in terms of ## grams ## of the weight of the atom. Why they defined an intermediate weight of ##amu##?
 

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