An atomic solid crystalliuzen in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by `60.3 "pm"` .If the atomic mass of A is 48, then density of the solid , is nearly :

Options

A. `2.7 g//"cc"`
B. `50.7 g//"cc"`
C. `3.5 g//"cc"`
D. `1.75 g//"cc"`

Answer

To solve the problem of finding the density of an atomic solid crystallized in a body-centered cubic (BCC) lattice, we will follow these steps:

### Step 1: Understand the Structure
In a body-centered cubic lattice, there are atoms located at each of the eight corners of the cube and one atom at the center of the cube. The total number of atoms per unit cell (z) can be calculated as follows:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell (since it is shared among 8 unit cells).
- There are 8 corner atoms, contributing \( 8 \times \frac{1}{8} = 1 \) atom.
- The center atom contributes 1 atom.
- Therefore, the total number of atoms per unit cell \( z = 1 + 1 = 2 \).

### Step 2: Relate the Edge Length to Atomic Radius
We know that the inner surface of the atoms at adjacent corners is separated by 60.3 pm. This distance is equal to the edge length \( a \) minus twice the atomic radius \( r \):
\[ a - 2r = 60.3 \text{ pm} \]
Thus, we can express the edge length \( a \) in terms of \( r \):
\[ a = 60.3 \text{ pm} + 2r \]

### Step 3: Use the Body Diagonal to Relate Edge Length and Radius
In a body-centered cubic lattice, the body diagonal can be expressed as:
\[ \text{Body Diagonal} = \sqrt{3}a \]
This diagonal also equals \( 4r \) (since it spans from one corner atom to the opposite corner atom through the center atom):
\[ \sqrt{3}a = 4r \]
From this, we can express \( r \) in terms of \( a \):
\[ r = \frac{\sqrt{3}}{4}a \]

### Step 4: Substitute \( r \) into the Edge Length Equation
Substituting \( r \) into the edge length equation:
\[ a = 60.3 \text{ pm} + 2\left(\frac{\sqrt{3}}{4}a\right) \]
This simplifies to:
\[ a = 60.3 \text{ pm} + \frac{\sqrt{3}}{2}a \]
Rearranging gives:
\[ a - \frac{\sqrt{3}}{2}a = 60.3 \text{ pm} \]
\[ a\left(1 - \frac{\sqrt{3}}{2}\right) = 60.3 \text{ pm} \]
Calculating \( 1 - \frac{\sqrt{3}}{2} \approx 0.134 \):
\[ a \cdot 0.134 = 60.3 \text{ pm} \]
\[ a = \frac{60.3 \text{ pm}}{0.134} \approx 450 \text{ pm} \]

### Step 5: Calculate the Density
Now we can calculate the density using the formula:
\[ \text{Density} = \frac{z \cdot m}{N_A \cdot a^3} \]
Where:
- \( z = 2 \) (number of atoms per unit cell)
- \( m = 48 \text{ g/mol} \) (atomic mass of A)
- \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \) (Avogadro's number)
- Convert \( a \) from pm to cm: \( 450 \text{ pm} = 450 \times 10^{-12} \text{ m} = 4.5 \times 10^{-8} \text{ cm} \)

Now substituting the values:
\[ \text{Density} = \frac{2 \cdot 48}{6.022 \times 10^{23} \cdot (4.5 \times 10^{-8})^3} \]
Calculating \( (4.5 \times 10^{-8})^3 \):
\[ (4.5 \times 10^{-8})^3 = 9.1125 \times 10^{-24} \text{ cm}^3 \]
Now substituting this value into the density formula:
\[ \text{Density} = \frac{96}{6.022 \times 10^{23} \cdot 9.1125 \times 10^{-24}} \approx 1.75 \text{ g/cm}^3 \]

### Final Answer
The density of the solid is approximately **1.75 g/cm³**.

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