The density of nickel (face centered cubic cell) is `8.94g//cm^(3) at 20^(@)C`. What is the radius of the atom? `("Atomic mass": Ni=59)`

Options

A. 0.124nm
B. 0.136nm
C. 0.149nm
D. 0.110nm

Answer

To find the radius of the nickel atom in a face-centered cubic (FCC) cell given its density and atomic mass, we can follow these steps:

### Step 1: Understand the FCC structure
In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell (Z = 4). The relationship between the edge length (a) of the cube and the radius (r) of the atom is given by:
\[ a = \frac{4r}{\sqrt{2}} \]
This can be rearranged to express r in terms of a:
\[ r = \frac{a \sqrt{2}}{4} \]

### Step 2: Use the formula for density
The density (d) of a substance can be expressed in terms of its mass (m), volume (V), and number of atoms (Z) in the unit cell:
\[ d = \frac{Z \cdot M}{N_A \cdot a^3} \]
Where:
- \( d \) = density (g/cm³)
- \( Z \) = number of atoms per unit cell (4 for FCC)
- \( M \) = molar mass (g/mol)
- \( N_A \) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \( a \) = edge length of the unit cell (cm)

### Step 3: Substitute known values
Given:
- Density \( d = 8.94 \, \text{g/cm}^3 \)
- Molar mass of Ni \( M = 59 \, \text{g/mol} \)
- \( Z = 4 \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Substituting these values into the density formula:
\[ 8.94 = \frac{4 \cdot 59}{6.022 \times 10^{23} \cdot a^3} \]

### Step 4: Solve for a³
Rearranging the equation to solve for \( a^3 \):
\[ a^3 = \frac{4 \cdot 59}{8.94 \cdot 6.022 \times 10^{23}} \]

Calculating the right-hand side:
\[ a^3 = \frac{236}{8.94 \cdot 6.022 \times 10^{23}} \]
\[ a^3 \approx \frac{236}{5.375 \times 10^{24}} \]
\[ a^3 \approx 4.39 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Calculate a
Taking the cube root to find \( a \):
\[ a \approx (4.39 \times 10^{-23})^{1/3} \]
\[ a \approx 3.48 \times 10^{-8} \, \text{cm} \]

### Step 6: Calculate the radius r
Now, substituting \( a \) back into the equation for \( r \):
\[ r = \frac{a \sqrt{2}}{4} \]
\[ r \approx \frac{3.48 \times 10^{-8} \cdot \sqrt{2}}{4} \]
\[ r \approx \frac{3.48 \times 10^{-8} \cdot 1.414}{4} \]
\[ r \approx \frac{4.92 \times 10^{-8}}{4} \]
\[ r \approx 1.23 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert to nanometers
To convert to nanometers:
\[ r \approx 1.23 \times 10^{-8} \, \text{cm} = 1.23 \times 10^{-6} \, \text{m} = 12.3 \, \text{nm} \]

### Final Answer
The radius of the nickel atom is approximately \( 1.23 \, \text{nm} \).

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