iron has a density of 7.86 g/cm3 and crystallizes a body centered cubic lattice. Show that only 68% of a body centered lattice is actually occupied by atoms, and determine the atomic radius of iron
bcc is 2 atoms/unit cell.
mass unit cell = 2*55.847/6.02E23 = ?
volume unit cell = mass unit cell/density
a = edge length = volume1/3
4r = a*(3)1/2 and solve for r = approximately 1.2E-8 cm.
volume of atom = (4/3)*pi*r3
ratio is space used =
(2*volume of atom/volume of unit cell)*100 = 68%
Well, iron seems to be quite a popular choice among atoms for their little party in the lattice. But despite their enthusiasm, they aren't able to occupy the entire lattice, leaving a bit of a vacuum in there. How rude, right?
To calculate the percentage of the lattice occupied by atoms, we need to consider the arrangement of the atoms in a body centered cubic lattice. Here's the joke: the atoms in this type of lattice are like a typical party scenario – they occupy the corners and the center of the room.
In a body centered cubic lattice, each corner atom contributes 1/8th of itself to the unit cell, whereas the center atom contributes the whole shebang. So, we have one full atom contributed by the center and 8 corner atoms, adding up to a total of 9/8th of an atom in each unit cell.
But hey, remember, there's only space for one atom in each unit cell. So, the 9/8th has to be slimmed down to fit the 1-atom limit. Using a little math magic, we can calculate that approximately 68% (1 / 9/8) of the body centered cubic lattice is actually occupied by atoms. Party space logistics, I tell ya!
Now, let's move on to determining the atomic radius of iron. To do that, we can make use of the given density and the formula relating density to atomic radius, which is:
Density = (4 × Atomic Mass) / (π × Atomic Radius^3)
Plugging in the density of iron (7.86 g/cm³), we can rearrange the formula to solve for the atomic radius, like juggling numbers on a unicycle. After putting in the values, we get the atomic radius of iron as approximately 1.24 Å (angstroms), or 124 picometers (pm). That's smaller than a clown shoe for sure!
So, to recap: only 68% of the body centered lattice is actually occupied by atoms, and the atomic radius of iron is about 1.24 Å or 124 pm. Now you know the dimensions of this atomic party, my friend!
To show that only 68% of a body-centered lattice is occupied by atoms, we need to calculate the packing efficiency of the lattice.
1. We know that the body-centered cubic (bcc) lattice has 2 atoms per unit cell: one at each of the 8 corner positions and one at the center.
2. The volume of a unit cell in a bcc lattice can be calculated using the formula: V = a^3, where "a" is the length of the edge of the unit cell.
3. Given that the density of iron is 7.86 g/cm^3, we can convert this to kg/m^3 by multiplying by 1000 (since there are 1000 cm in 1 meter): density = 7.86 x 1000 = 7860 kg/m^3.
4. We know that the molar mass of iron (Fe) is approximately 55.845 g/mol.
5. Using Avogadro's number (6.022 x 10^23 atoms/mol), we can calculate the volume occupied by a single iron atom (Vatom) using the formula: Vatom = (Molar mass of Fe / Density of Fe) x (1 mol/Avogadro's number).
6. Once we have Vatom, we can calculate the volume fraction of the unit cell that is occupied by iron atoms.
7. Finally, we can calculate the atomic radius (r) using the formula: r = (3Vatom / (4π))^(1/3).
Let's solve the problem step-by-step:
Step 1: Calculate the volume occupied by a single iron atom (Vatom):
Vatom = (Molar mass of Fe / Density of Fe) x (1 mol/Avogadro's number)
Vatom = (55.845 g/mol / 7860 kg/m^3) x (1 mol / 6.022 x 10^23 atoms)
(Note: Remember to convert grams to kilograms before performing the calculation.)
Step 2: Calculate the volume of the unit cell (V):
V = a^3
(Note: Since the bcc lattice has a body-centered atom, the length of the edge of the unit cell (a) is equal to 4 times the atomic radius (r).)
Step 3: Calculate the volume fraction of the unit cell occupied by iron atoms:
Volume fraction = (Number of atoms x Vatom) / V
(Note: The number of atoms in a bcc unit cell is 2.)
Step 4: Calculate the atomic radius (r):
r = (3Vatom / (4π))^(1/3)
Start by calculating Vatom in step 1, and continue the calculations in the subsequent steps.
To calculate the percentage of a body-centered cubic (BCC) lattice that is occupied by atoms, we need to consider the arrangement of atoms in the crystal structure.
In a BCC lattice, each corner of the unit cell contains one atom, and there is one additional atom at the center of the unit cell. So, a BCC unit cell consists of 2 atoms.
To determine the percentage of the lattice occupied by atoms, we need to divide the volume occupied by the atoms by the total volume of the unit cell.
To calculate the volume occupied by the atoms, we can consider that each atom is a sphere. The volume of a sphere can be calculated using the formula:
Volume = (4/3) * pi * radius^3
Let's assume the atomic radius of iron is 'r'.
For the occupied volume, we have two atoms, one at the corner and one at the center. So the volume occupied by the atoms in one unit cell is:
Volume occupied = 2 * [(4/3) * pi * r^3]
The total volume of the BCC unit cell can be determined by multiplying the edge length of the unit cell cubed:
Total volume = (edge length)^3
Now, we know that the edge length of the BCC unit cell can be determined using the formula:
Edge length = 4 * radius / sqrt(3)
Substituting this expression for the edge length into the equation for the total volume:
Total volume = [(4 * radius / sqrt(3))]^3
Now, let's substitute the volume occupied by atoms and total volume calculations into the expression for the occupied percentage:
Occupied percentage = (Volume occupied / Total volume) * 100
Substituting the formulas and simplifying the expression:
Occupied percentage = [2 * (4/3) * pi * r^3] / [(4 * r / sqrt(3))^3] * 100
Occupied percentage = [2 * (4/3) * pi * r^3] / [(64 * r^3) / (27)] * 100
Occupied percentage = [(8/3) * pi * r^3] / [(64/27) * r^3] * 100
Further simplifying the expression:
Occupied percentage = (8/3) * (27/64) * 100
Occupied percentage = 68.75%
Therefore, we have found that approximately 68% of the body-centered cubic lattice is occupied by atoms.
To determine the atomic radius of iron, we can rearrange the equation for the edge length:
Edge length = 4 * radius / sqrt(3)
Solving for the atomic radius:
Atomic radius = (Edge length * sqrt(3)) / 4
Given that the density of iron is 7.86 g/cm3, we can calculate the edge length using the equation:
Edge length = (3 * molar mass) / (6.022 x 10^23 * density)
The molar mass of iron (Fe) is approximately 55.85 g/mol.
Substituting the values and calculating:
Edge length = (3 * 55.85 g/mol) / (6.022 x 10^23 * 7.86 g/cm3)
Edge length = (167.55 / 4.714) x 10^-23 cm
Edge length ≈ 3.525 x 10^-8 cm
Now, we can substitute the edge length into the equation for the atomic radius:
Atomic radius = (3.525 x 10^-8 cm * sqrt(3)) / 4
Atomic radius ≈ 1.923 x 10^-8 cm
Therefore, the atomic radius of iron is approximately 1.923 x 10^-8 cm.