If the radius of metal atom is 1.00 Å and its crystal structure is simple cubic, then what is the volume of one unit cell?

Options

A. `8xx10^(-28)m^3`
B. `4xx10^(-30)cm^3`
C. `8xx10^(-20)m^3`
D. `2xx10^(-24)cm^3`

Explanation

To find the volume of one unit cell of a simple cubic crystal structure given the radius of a metal atom, we can follow these steps:

### Step 1: Understand the Simple Cubic Structure
In a simple cubic structure, atoms are located at the corners of the cube. Each corner atom is shared by eight adjacent cubes, meaning that each cube effectively contains 1 atom (since there are 8 corners and each corner contributes 1/8 of an atom).

### Step 2: Relate the Edge Length (A) to the Radius (R)
In a simple cubic lattice, the atoms at the corners touch each other along the edge of the cube. Therefore, the relationship between the edge length (A) of the cube and the radius (R) of the atom is given by:
\[ A = 2R \]

### Step 3: Substitute the Given Radius
Given that the radius \( R \) of the metal atom is 1.00 Å (angstrom), we convert this to meters:
\[ R = 1.00 \, \text{Å} = 1.00 \times 10^{-10} \, \text{m} \]

Now, substituting this value into the equation for A:
\[ A = 2R = 2 \times (1.00 \times 10^{-10} \, \text{m}) = 2.00 \times 10^{-10} \, \text{m} \]

### Step 4: Calculate the Volume of the Unit Cell
The volume \( V \) of a cube is given by the formula:
\[ V = A^3 \]

Substituting the value of A we found:
\[ V = (2.00 \times 10^{-10} \, \text{m})^3 \]
\[ V = 8.00 \times 10^{-30} \, \text{m}^3 \]

### Step 5: Convert the Volume to Centimeters Cubed
To convert from cubic meters to cubic centimeters, we use the conversion factor \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \):
\[ V = 8.00 \times 10^{-30} \, \text{m}^3 \times 10^6 \, \text{cm}^3/\text{m}^3 \]
\[ V = 8.00 \times 10^{-24} \, \text{cm}^3 \]

### Final Answer
The volume of one unit cell is:
\[ V = 8.00 \times 10^{-30} \, \text{m}^3 \text{ or } 8.00 \times 10^{-24} \, \text{cm}^3 \]

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