Calculation of Curie Constant for Iron in Metallic Form

Click For Summary

Homework Help Overview

The problem involves calculating the Curie constant for iron in its metallic form, given specific parameters such as the Curie temperature, magnetic moment, and lattice structure. The subject area pertains to magnetism and solid-state physics.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the Curie constant using a formula that incorporates magnetic moment, number of atoms per unit volume, and other constants. Some participants question the assumption regarding the number of atoms per lattice cube in a body-centered cubic structure.

Discussion Status

The discussion is ongoing, with participants exploring the validity of the assumptions made in the calculations. There is no explicit consensus on the correct number of atoms per lattice cube, and some guidance has been suggested regarding checking atomic packing factors.

Contextual Notes

The original poster expresses frustration over a lack of responses and seeks clarification on the calculations. There is an indication of a discrepancy between the calculated and expected values for the Curie constant.

Amith2006
Messages
416
Reaction score
2

Homework Statement



As none are responding to my query posted in Introductory Physics section, I am posting it here. Someone please guide me!
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.


Homework Equations


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



The Attempt at a Solution



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^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.
 
Physics news on Phys.org
I suspect the difference is related to the phrase "body centered cube". Are you sure there are 2 atoms per lattice cube? I'd check around under subjects like "atomic packing factor" to make sure.
 
On second thought the 2 seems ok. Hmm. I'm no expert in this field.
 
What on Earth is preventing you all from answering?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
10K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
7K