How Do You Determine Atomic Distance from Density and Molar Mass?

  • Thread starter Thread starter Azzy42
  • Start date Start date
  • Tags Tags
    Modulus
Click For Summary
SUMMARY

The discussion focuses on determining atomic distance in lead using its density and molar mass. Key values include Young's modulus for lead at 1.6 x 1010 N/m2, a density of 11.4 g/cm3, and a molar mass of 207 g/mol. The relationship between Young's modulus (Y), spring constant (K), and atomic distance (d) is established through the formula Y = K/d. The challenge lies in calculating the atomic diameter (d) from the given density and molar mass.

PREREQUISITES
  • Understanding of Young's modulus and its application in material science.
  • Knowledge of density calculations and conversions between units (g/cm3 to kg/m3).
  • Familiarity with molar mass and its significance in atomic calculations.
  • Basic principles of atomic theory, particularly regarding atomic models and point masses.
NEXT STEPS
  • Research how to calculate atomic volume from density and molar mass.
  • Learn about the relationship between atomic distance and atomic radius in crystalline structures.
  • Explore the concept of spring constants in atomic interactions and their derivation.
  • Investigate the implications of treating atoms as point masses in physical models.
USEFUL FOR

Students and professionals in materials science, physicists, and chemists interested in atomic structure and properties of lead.

Azzy42
Messages
2
Reaction score
0
Heres my problem.
Youngs modulus for lead: 1.6exp10 N/m^2. Density of Lead: 11,4g/cm^3.
Mass of one mole lead: 207g
Model the interactions as if they were connected by springs.
Determine distance between lead atoms and approx spring constant.

Easy enough Y= K/d But what is bugging me is d. dia of atom
How do you determine d from density and the mass of one mole?
And how do u relate it to the distance between atoms?
It is keeping me up.
Probably very easy but just not seeing it.

PS: don't know if we use different notation is SA
 
Physics news on Phys.org
If you assume that each atom of lead is a point mass (has mass but zero volume), then the problem is greatly simplified.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
7K
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
21K
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K