By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be ` 19.4 g cm^(-3)` and its atomic mass is 197 a.m.u.
Assuming gold atom to be spherical , its radius will be

Options

A. 203.5 "pm"^(2)`
B. 143.9 pm
C. 176. 2 pm
D. 287 . 8 pm

Explanation

To find the radius of a gold atom given its density and atomic mass, we can follow these steps:

### Step 1: Understand the relationship between density, atomic mass, and volume
The density (\( \rho \)) of a substance is given by the formula:
\[ \rho = \frac{Z \cdot M}{A \cdot V} \]
Where:
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass (in grams per mole)
- \( A \) = Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mole)
- \( V \) = volume of the unit cell (in cm³)

### Step 2: Calculate the volume of the unit cell
Rearranging the formula gives us:
\[ V = \frac{Z \cdot M}{\rho \cdot A} \]
For gold, we have:
- \( Z = 4 \) (since every fourth layer is identical)
- \( M = 197 \, \text{g/mol} \)
- \( \rho = 19.4 \, \text{g/cm}^3 \)
- \( A = 6.022 \times 10^{23} \, \text{atoms/mol} \)

Substituting these values into the equation:
\[ V = \frac{4 \cdot 197}{19.4 \cdot 6.022 \times 10^{23}} \]

### Step 3: Calculate the volume
Calculating the numerator:
\[ 4 \cdot 197 = 788 \]
Calculating the denominator:
\[ 19.4 \cdot 6.022 \times 10^{23} \approx 1.167 \times 10^{24} \]
Now substituting these values:
\[ V = \frac{788}{1.167 \times 10^{24}} \approx 6.75 \times 10^{-22} \, \text{cm}^3 \]

### Step 4: Calculate the edge length of the unit cell
The volume of a cubic unit cell is given by:
\[ V = a^3 \]
Where \( a \) is the edge length. Thus:
\[ a = V^{1/3} = (6.75 \times 10^{-22})^{1/3} \]
Calculating \( a \):
\[ a \approx 4.07 \times 10^{-8} \, \text{cm} = 407 \, \text{pm} \, (\text{picometers}) \]

### Step 5: Calculate the radius of the gold atom
For a face-centered cubic (FCC) structure, the relationship between the radius \( r \) and the edge length \( a \) is given by:
\[ r = \frac{a}{2\sqrt{2}} \]
Substituting the value of \( a \):
\[ r = \frac{407}{2\sqrt{2}} \approx 143.9 \, \text{pm} \]

### Final Answer
The radius of a gold atom is approximately \( 143.9 \, \text{pm} \).

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