I have a problem set due tomorrow, and two problems in it were not covered in class.

The first is this:
Molybdenum has an atomic radius of 0.145 nm. The volume of its cubic unit cell is 0.0375 nm^3. What is the geometry of the molybdenum unit cell?

I have no idea where to even begin with this problem.

There may be an easier way to do this but here is what I do.
for bcc(body centered cubic)
radius =a(sqrt 3)/4

for face centered cubic
radius = a(sqrt 2)/4

for simple cubic
radius = a/2

so you find the one that fits your data. I know Mo is a bcc; therefore, we try that. First, a=V1/3
V=0.0375; therefore,
a=(0.03751/3, then
0.03751/3*1.732/4= 0.145 and that is the radius given in the problem. sqrt 3 = 1.732

Try that for the others and radius won't be 0.145.
I hope this helps a little

Bot Bot answered
1 year ago

.

Therefore, the geometry of the molybdenum unit cell is body-centered cubic (bcc).

Step-by-Step Bot Step-by-Step Bot answered
8 months ago

To determine the geometry of the molybdenum unit cell, we can follow these steps:

1. Recall the formulas for calculating the radius of a unit cell for different geometries:
- For a Body-Centered Cubic (BCC) unit cell: radius = a * (√3) / 4
- For a Face-Centered Cubic (FCC) unit cell: radius = a * (√2) / 4
- For a Simple Cubic unit cell: radius = a / 2

2. Since we are given the atomic radius of molybdenum (0.145 nm), we can compare it to the calculated radii for each unit cell geometry.

3. Start with the BCC unit cell since molybdenum is known to have a BCC structure. Let's calculate the lattice constant 'a':
- The volume of a cubic unit cell is given as 0.0375 nm^3.
- We can find the lattice constant 'a' by taking the cube root of the volume: a = (0.0375 nm^3)^(1/3).
- Calculate 'a' to get the lattice constant.

4. Now, substitute the calculated lattice constant 'a' into the formula for the BCC unit cell radius:
radius = a * (√3) / 4.
- Calculate the radius using the given lattice constant 'a'.

5. Compare the calculated radius to the given atomic radius (0.145 nm).
- If the calculated radius matches the given atomic radius, then the molybdenum unit cell has a BCC geometry.
- If the radius does not match, then you can try calculating the radii for FCC or Simple Cubic unit cell geometries using the same steps as above.

Remember to use the appropriate units consistently throughout the calculations.

Explain Bot Explain Bot answered
8 months ago

To solve this problem, we need to determine the geometry of the molybdenum unit cell based on the given atomic radius and volume of the cubic unit cell.

There are three types of cubic unit cells: simple cubic, body-centered cubic (bcc), and face-centered cubic (fcc).

We can start by considering the simple cubic structure. In this case, the atomic radius (r) is equal to half the length of the unit cell's edge (a).

For the bcc structure, the atomic radius is related to the edge length (a) by the formula:

radius = a * sqrt(3) / 4

Similarly, for the fcc structure, the atomic radius is related to the edge length (a) by the formula:

radius = a * sqrt(2) / 4

By comparing the value of the given atomic radius (0.145 nm) to the calculated values using these formulas, we can determine which structure (simple cubic, bcc, or fcc) matches the given data.

Let's now use these formulas to determine the geometry of the molybdenum unit cell:

1. For the simple cubic structure:
Radius = a / 2

2. For the bcc structure:
Radius = a * sqrt(3) / 4

3. For the fcc structure:
Radius = a * sqrt(2) / 4

Using the given data, we have the atomic radius as 0.145 nm and the volume of the cubic unit cell as 0.0375 nm^3.

Let's first determine the edge length (a) using the volume of the unit cell (V = a^3):

a = V^(1/3)
a = (0.0375 nm^3)^(1/3)
a ≈ 0.343 nm

Now, let's substitute this value of a into the formulas for the different structures to calculate the corresponding radii:

1. For simple cubic:
Radius = a / 2
Radius ≈ 0.343 nm / 2
Radius ≈ 0.172 nm

2. For bcc:
Radius = a * sqrt(3) / 4
Radius ≈ 0.343 nm * sqrt(3) / 4
Radius ≈ 0.297 nm

3. For fcc:
Radius = a * sqrt(2) / 4
Radius ≈ 0.343 nm * sqrt(2) / 4
Radius ≈ 0.242 nm

By comparing the calculated radii with the given atomic radius (0.145 nm), we see that the closest match is the bcc structure with a radius of approximately 0.297 nm.

Therefore, the geometry of the molybdenum unit cell is body-centered cubic (bcc).

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