Tungsten crystallises in a body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom ?

Explanation

To find the radius of a tungsten atom in a body-centered cubic (BCC) unit cell, we can follow these steps:

### Step 1: Understand the structure of the BCC unit cell
In a body-centered cubic unit cell, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube. The atoms at the corners are in contact with the atom at the center along the body diagonal.

### Step 2: Identify the relationship between the edge length and the radius
The body diagonal of the cube can be expressed in terms of the edge length (a) of the cube. The formula for the body diagonal (d) in a cubic cell is:
\[ d = \sqrt{3}a \]

### Step 3: Relate the body diagonal to the atomic radii
In a BCC structure, the body diagonal is equal to the sum of the diameters of the atoms along that diagonal. Since there are two corner atoms (each contributing a radius) and one body-centered atom (contributing a diameter), we can express this as:
\[ d = 4r \]
where \( r \) is the radius of the tungsten atom.

### Step 4: Set up the equation
From the above relationships, we can equate the two expressions for the body diagonal:
\[ \sqrt{3}a = 4r \]

### Step 5: Solve for the radius \( r \)
Rearranging the equation gives:
\[ r = \frac{\sqrt{3}a}{4} \]

### Step 6: Substitute the given edge length
Now, substitute the given edge length \( a = 316.5 \, \text{pm} \):
\[ r = \frac{\sqrt{3} \times 316.5 \, \text{pm}}{4} \]

### Step 7: Calculate the value
First, calculate \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
Now substitute this value:
\[ r = \frac{1.732 \times 316.5}{4} \]

Calculating this gives:
\[ r = \frac{548.298}{4} \approx 137.07 \, \text{pm} \]

### Final Answer
The radius of the tungsten atom is approximately \( 137 \, \text{pm} \).

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