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Calculation of curie constant of iron

  • Thread starter Amith2006
  • Start date
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2
1. Homework Statement
1)The curie temperature of iron is 1043 Kelvin. Assume that iron atoms, when in metallic form have moments of 2 Bohr magneton per atom. Iron is body centered cube with lattice parameter a = 0.286 nm. Calculate the curie constant.



2. Homework Equations

C = [(m^2)(mu)N]/[3K]


3. The Attempt at a Solution

I solved it in the following way:
Let m be the magnetic moment of an iron atom, let N be the number of atoms per unit volume, let K be the boltzmann constant, let mu be the permeability of free space and let C be the Curie constant.
m = 2[m(B)] {where m(B) is Bohr magneton}
= 18.54 x 10^(-24) A-m^2
N = n/(a^3) {where n is number of atoms in 1 cubic lattice of iron}
= 2/[(0.286 x 10^(-9))^3]
= 8.5 x 10^28 atoms per unit volume
C = [(m^2)(mu)N]/[3K]
C = 0.89
But the answer given in my book is 0.66. Please help!
 

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