In the crystalline solids the smallest repeating part in the lattice is known as unit cell. The unit cells are described as simple (points at all corners), body centred (points at all the corners and it the centre), face centred (points at all the corners and centre of all faces), and end centred (points at all hte corners and centres of two opposite and faces) unit cells. In two common tyupes of packing ccp and hcp, 26% of space is left unoccupied in the form of interstitial sites. For the stable ionic crystalline structures, there is difinite radius ratio limit for a cation to fit perfectly in the lattice of anions, called radius ratio rule. This also defines the coordination number of an ion, which is the number of nearest neighbours of opposite charges. This depeds upon the ratio of radii of two types of ions, `r_(+)//r_(-)`. This ratio for coordination numbers 3,4,6,and 8 is respectively 0.155 - 0.225, 0.225 - 0.414, 0.414 - 0.732 and 0.732 - 1 respectively.
Gold crystallizes in a face centred unit cell. Its edge length is 0.410 nm. The radius of gold atom is

Options

A. 0.205 nm
B. 0.290 nm
C. 0.145 nm
D. 0.578 nm

Explanation

To find the radius of a gold atom in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step 1: Understand the structure of FCC
In a face-centered cubic (FCC) unit cell, atoms are located at each corner of the cube and at the center of each face. This means that there are 8 corner atoms and 6 face atoms contributing to the unit cell.

### Step 2: Identify the relationship between edge length and atomic radius
In an FCC unit cell, the face diagonal can be expressed in terms of the atomic radius (r) and the edge length (a) of the unit cell. The face diagonal is equal to 4 times the atomic radius because it passes through the center of one face atom and touches two corner atoms.

### Step 3: Calculate the face diagonal
Using the Pythagorean theorem, the face diagonal (d) can be calculated as:
\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 4: Set up the equation for the face diagonal
Since the face diagonal is also equal to 4 times the atomic radius (r), we can set up the equation:
\[ d = 4r \]
Thus, we can equate the two expressions:
\[ a\sqrt{2} = 4r \]

### Step 5: Solve for the atomic radius (r)
Rearranging the equation gives us:
\[ r = \frac{a\sqrt{2}}{4} \]

### Step 6: Substitute the given edge length
We are given that the edge length (a) of gold is 0.410 nm. Now, we can substitute this value into the equation:
\[ r = \frac{0.410 \, \text{nm} \cdot \sqrt{2}}{4} \]

### Step 7: Calculate the radius
First, calculate \(\sqrt{2}\):
\[ \sqrt{2} \approx 1.414 \]
Now substitute this back into the equation:
\[ r = \frac{0.410 \cdot 1.414}{4} \]
\[ r = \frac{0.57974}{4} \]
\[ r \approx 0.144935 \, \text{nm} \]

### Step 8: Round the result
Rounding this to three significant figures gives:
\[ r \approx 0.145 \, \text{nm} \]

### Final Answer
The radius of the gold atom is approximately **0.145 nm**.

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