In the closed packing of atoms A `("radius":r_(a))`, the radius of atom B that can be fitted into tetrahedral void is:

Options

A. `0.155r_(a)`
B. `0.255r_(a)`
C. `0.414r_(a)`
D. `0.732r_(a)`

Answer

To find the radius of atom B that can be fitted into the tetrahedral void formed by atoms A in a closed packing arrangement, we can follow these steps:

### Step 1: Understand the arrangement of atoms
In a closed packing of atoms A, the atoms are arranged in a way that maximizes the space occupied. The tetrahedral void is formed by four atoms A, which can be visualized as occupying the corners of a tetrahedron.

### Step 2: Define the radius of atom A
Let the radius of atom A be denoted as \( r_A \).

### Step 3: Calculate the edge length of the unit cell
In a close-packed structure, the edge length \( a \) of the unit cell can be related to the radius of the atoms. For a face-centered cubic (FCC) arrangement, the relationship is:
\[ a = 2\sqrt{2} r_A \]

### Step 4: Calculate the face diagonal
The face diagonal \( d_f \) of the cube can be calculated using the Pythagorean theorem:
\[ d_f = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 5: Relate the face diagonal to the radius of atom A
The face diagonal will be equal to the sum of the diameters of two atoms A:
\[ d_f = 2r_A + 2r_A = 4r_A \]
Thus, we have:
\[ a\sqrt{2} = 4r_A \]

### Step 6: Calculate the body diagonal
The body diagonal \( d_b \) of the cube can be calculated as:
\[ d_b = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} \]

### Step 7: Relate the body diagonal to the radii of atoms A and B
The body diagonal will contain two atoms A and one atom B:
\[ d_b = 2r_A + 2r_B \]
Thus, we have:
\[ a\sqrt{3} = 2r_A + 2r_B \]

### Step 8: Set up the equations
Now we have two equations:
1. \( a\sqrt{2} = 4r_A \)
2. \( a\sqrt{3} = 2r_A + 2r_B \)

### Step 9: Solve for \( r_B \)
From the first equation, we can express \( a \) in terms of \( r_A \):
\[ a = \frac{4r_A}{\sqrt{2}} = 2\sqrt{2}r_A \]

Substituting this value of \( a \) into the second equation gives:
\[ 2\sqrt{2}r_A\sqrt{3} = 2r_A + 2r_B \]

### Step 10: Simplify and solve for \( r_B \)
Dividing the entire equation by 2:
\[ \sqrt{6}r_A = r_A + r_B \]

Rearranging gives:
\[ r_B = \sqrt{6}r_A - r_A = (\sqrt{6} - 1)r_A \]

### Step 11: Calculate the numerical value
Using the approximate value of \( \sqrt{6} \approx 2.45 \):
\[ r_B \approx (2.45 - 1)r_A \approx 1.45r_A \]

### Conclusion
The radius of atom B that can be fitted into the tetrahedral void is approximately \( 0.225r_A \).