If a metal crystallises in bcc structure with edge length of unit cell `4.29 xx 10^(-8) cm`, the radius of metal atom is

Options

A. `1.07 xx 10^(-7) cm`
B. `1.07 xx 10^(-8) cm`
C. `1.86 xx 10^(-8) cm`
D. `3.2 xx 10^(-7) cm`

Explanation

To find the radius of a metal atom that crystallizes in a body-centered cubic (BCC) structure with a given edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**: In a BCC structure, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube.

2. **Understand the Geometry**: The atoms at the corners and the center touch along the body diagonal of the cube.

3. **Calculate the Body Diagonal**: The body diagonal (d) of a cube can be calculated using the formula:
\[ d = \sqrt{3} \times a \]
where \( a \) is the edge length of the cube.

4. **Relate the Body Diagonal to Atomic Radius**: In a BCC structure, the body diagonal is equal to four times the radius of the atom (since there are two radii from the corner atom to the center atom):
\[ d = 4r \]

5. **Combine the Equations**: By substituting the expression for the body diagonal into the equation relating it to the radius, we get:
\[ \sqrt{3} \times a = 4r \]

6. **Solve for the Radius**: Rearranging the equation gives:
\[ r = \frac{\sqrt{3}}{4} \times a \]

7. **Substitute the Given Edge Length**: Now, substitute the given edge length \( a = 4.29 \times 10^{-8} \) cm into the equation:
\[ r = \frac{\sqrt{3}}{4} \times 4.29 \times 10^{-8} \text{ cm} \]

8. **Calculate the Radius**: Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\[ r = \frac{1.732}{4} \times 4.29 \times 10^{-8} \text{ cm} \]
\[ r = 0.433 \times 4.29 \times 10^{-8} \text{ cm} \]
\[ r \approx 1.86 \times 10^{-8} \text{ cm} \]

### Final Answer:
The radius of the metal atom is approximately \( r \approx 1.86 \times 10^{-8} \) cm.

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