In a body-centred cubic crystal of an element , the ratio of the readius of the radius of the atom toi the edge of the unit cell is ……

Answer

To find the ratio of the radius of the atom (r) to the edge length of the unit cell (a) in a body-centered cubic (BCC) crystal structure, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic (BCC) structure, there is one atom at each of the eight corners of the cube and one atom at the center of the cube.

### Step 2: Analyze the Geometry
In a BCC unit cell, the atoms at the corners touch the atom at the center along the body diagonal of the cube. The body diagonal can be calculated using the Pythagorean theorem.

### Step 3: Calculate the Body Diagonal
The length of the body diagonal (d) of a cube with edge length (a) can be calculated as:
\[ d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

### Step 4: Relate the Body Diagonal to Atomic Radius
In the BCC structure, the body diagonal is equal to four times the radius of the atom (4r) because there are two radii from the corner atom to the center atom and two more radii from the center atom to the other corner atom:
\[ d = 4r \]

### Step 5: Set Up the Equation
Now we can set up the equation:
\[ a\sqrt{3} = 4r \]

### Step 6: Solve for the Ratio
To find the ratio of the radius of the atom to the edge length of the unit cell, we rearrange the equation:
\[ r = \frac{a\sqrt{3}}{4} \]

Now, to find the ratio \( \frac{r}{a} \):
\[ \frac{r}{a} = \frac{\sqrt{3}}{4} \]

### Final Answer
The ratio of the radius of the atom to the edge of the unit cell in a body-centered cubic crystal is:
\[ \frac{r}{a} = \frac{\sqrt{3}}{4} \]

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