Respuesta :
Answer:
The unit cell edge lenght in pm is equal to 361 pm
Explanation:
Data provided:
ρ=Copper density=8.96 g/cm3
Atomic mass of copper=63.54 g/mol
Atoms/cell=4 atoms (in theory)
Avogadro's number=6.02x[tex]10^{23}[/tex] atoms/mol
Since copper has a cubic structure, its cell volume is equal to [tex]a^{3}[/tex], which can be obtained through the relationship:
cell volume=[tex]\frac{(atoms/cell)(atomic mass)}{(density)(Avogadros number)}[/tex]
Substituting the values:
cell volume=[tex]\frac{(4 atoms)(63.54 g/mol)}{(8.96 g/cm3)(6.02x10^{23}) }=4.71x10^{-23}cm^{3}[/tex]
clearing, we have:
a=[tex]\sqrt[3]{4.71x10^{-23}cm^{3} }=3.61x10^{-8}cm[/tex]
We convert from centimeter to picometer, 1cm=1x[tex]10^{10}[/tex]pm
a=[tex]3.61x10^{-8}cmx\frac{1x10^{10}pm }{1cm} =361 pm[/tex]
The unit cell edge length will be "361 pm".
Given:
Density of copper,
- [tex]\rho = 8.96 \ g/cm^3[/tex]
Atomic mass,
- [tex]63.54 \ g/mol[/tex]
Atoms/cell,
- [tex]4 \ atoms[/tex]
Avogadro's number,
- [tex]6.02\times 10^{23} \ atms/mol[/tex]
As we know the formula,
→ [tex]Cell \ volume = \frac{(Atoms/cel) (Atomic \ mass)}{(Density)(Avogadro's \ number)}[/tex]
By substituting the values, we get
→ [tex]= \frac{4\times 63.54}{8.96\times 6.02\times 10^{23}}[/tex]
→ [tex]= 4.71\times 10^{-23} \ cm^3[/tex]
Now,
→ [tex]a = \sqrt[3]{4.71\times 10^{-23}}[/tex]
[tex]= 3.61\times 10^{-8} \ cm[/tex]
or,
[tex]=361 \ pm[/tex]
Thus the above answer is correct.
Learn more:
https://brainly.com/question/24262082
