Finding the number of free electrons

[PhysicsTest]

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?

 

[PeroK]

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 $kg$?

 

[PeroK]

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 $1g$.

 

[PhysicsTest]

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

 

[PeroK]

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 $d \ kg$

 

[PhysicsTest]

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 $amu$?