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# 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.I solved it in the following way: Let m be the magnetic moment of an iron atom, N be the number of atoms per unit volume, K be the boltzmann constant, mu be the permeability of free space and C be the Curie constant.m =  2[m(B)]        {where m(B) is Bohr magneton}            =  18.54 x 10^(-24) A-m^2N = 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  = 0.89But the answer given in my book is 0.66.

Grade:10

## 1 Answers

Abdullah Alvi
13 Points
one year ago
Hi Aditi!
According to my understanding you can use this formula and get the answer;
C=[N{(u0)^2}{u(B)^2}]/K
C=[{8.5*10^28}{4pi*(10^-7)}{(9.27*10^(-23))^2}]/[1.38*10^(-23)]
C=0.66
Hope you got it.
Thank you

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