Calc Lattice Constant of Solid Manganese

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SUMMARY

The discussion focuses on calculating the lattice constant of solid manganese, which has a simple cubic structure. Given the atomic mass (Ar) of 54.94 g/mol and a density of 7440 kg/m³, the calculation involves determining the number of atoms in the unit cell. The formula provided is a systematic approach to relate the number of atoms per cell, weight, and volume, ultimately leading to the determination of the lattice parameter.

PREREQUISITES
  • Understanding of simple cubic crystal structures
  • Familiarity with atomic mass and density concepts
  • Knowledge of unit cell volume calculations
  • Basic grasp of mole and Avogadro's number
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adambinch
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I am told solid manganese exists in a simple cubic structure. If its Ar is 54.94 and its density is 7440 Kg m^-3, How do I calculate the lattice constant for this material?
 
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Hi adambinch, welcome to PF. You need to find out how many atoms there are in the simple cubic unit cell. Then you can set up a calculation like

[tex]\left(\frac{\mathrm{atoms}}{\mathrm{cell}}\right)\left(\frac{\mathrm{weight}}{\mathrm{mole}}\right)\left(\frac{\mathrm{mole}}{\mathrm{atoms}}\right)\left(\frac{\mathrm{volume}}{\mathrm{weight}}\right)=\left(\frac{\mathrm{volume}}{\mathrm{cell}}\right)[/tex]

and the volume per unit cell should lead you to the lattice parameter. Does this help?
 

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