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 a nickel atom in a face-centered cubic (FCC) structure given the density and atomic mass, we can follow these steps:

### Step 1: Understand the formula for density in a cubic cell
The density (\(d\)) of a crystal structure can be calculated using the formula:
\[ d = \frac{Z \times M}{N_A \times a^3} \]
Where:
- \(Z\) = number of atoms per unit cell (for FCC, \(Z = 4\))
- \(M\) = molar mass of the substance (for Ni, \(M = 59 \, g/mol\))
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, mol^{-1}\))
- \(a\) = edge length of the cubic cell

### Step 2: Rearrange the formula to find \(a^3\)
Rearranging the formula to solve for \(a^3\):
\[ a^3 = \frac{Z \times M}{d \times N_A} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
- \(Z = 4\)
- \(M = 59 \, g/mol\)
- \(d = 8.94 \, g/cm^3\)
- \(N_A = 6.022 \times 10^{23} \, mol^{-1}\)

\[ a^3 = \frac{4 \times 59}{8.94 \times 6.022 \times 10^{23}} \]

### Step 4: Calculate \(a^3\)
Calculating the numerator:
\[ 4 \times 59 = 236 \]
Calculating the denominator:
\[ 8.94 \times 6.022 \times 10^{23} \approx 5.375 \times 10^{24} \]
Now substituting these values:
\[ a^3 = \frac{236}{5.375 \times 10^{24}} \approx 4.38484 \times 10^{-23} \, cm^3 \]

### Step 5: Calculate \(a\)
Now, take the cube root of \(a^3\) to find \(a\):
\[ a = (4.38484 \times 10^{-23})^{1/3} \approx 3.53 \times 10^{-8} \, cm \]

### Step 6: Relate edge length \(a\) to atomic radius \(r\)
In a face-centered cubic structure, the relationship between the edge length \(a\) and the atomic radius \(r\) is given by:
\[ a = 2\sqrt{2}r \]
Thus, we can solve for \(r\):
\[ r = \frac{a}{2\sqrt{2}} = \frac{3.53 \times 10^{-8}}{2\sqrt{2}} \approx 1.25 \times 10^{-8} \, cm \]

### Step 7: Convert \(r\) to nanometers
To convert \(r\) to nanometers:
\[ 1 \, cm = 10^{7} \, nm \implies r \approx 0.125 \, nm \]

### Final Answer
The radius of the nickel atom is approximately \(0.125 \, nm\).

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