`N_A` spheres of radius 'R' are melted in `x xxN_A` smaller spheres, so that when fcc lattice is generated from smaller spheres the edge length is formed to be `2 5/6 R`.Find x.

Explanation

To solve the problem, we need to determine the value of \( x \) given that \( N_A \) spheres of radius \( R \) are melted into \( x \times N_A \) smaller spheres and that the edge length of the FCC lattice formed from these smaller spheres is \( 2 \frac{5}{6} R \).

### Step-by-Step Solution:

1. **Volume of the Larger Spheres**:
The volume of one larger sphere of radius \( R \) is given by:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Therefore, the total volume of \( N_A \) larger spheres is:
\[ V_{\text{total large}} = N_A \times \frac{4}{3} \pi R^3 \]

2. **Volume of the Smaller Spheres**:
Let the radius of the smaller spheres be \( r \). The volume of one smaller sphere is:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of \( x \times N_A \) smaller spheres is:
\[ V_{\text{total small}} = x \times N_A \times \frac{4}{3} \pi r^3 \]

3. **Equating the Volumes**:
Since the total volume of the larger spheres is equal to the total volume of the smaller spheres, we can set the two expressions equal to each other:
\[ N_A \times \frac{4}{3} \pi R^3 = x \times N_A \times \frac{4}{3} \pi r^3 \]
Canceling \( N_A \) and \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = x r^3 \]

4. **Relating the Radii**:
In a face-centered cubic (FCC) lattice, the relationship between the edge length \( a \) and the radius \( r \) of the spheres is given by:
\[ a = 2\sqrt{2} r \]
We are given that the edge length \( a \) is \( 2 \frac{5}{6} R \), which can be expressed as:
\[ a = \frac{17}{6} R \]

5. **Setting Up the Equation**:
Substituting the expression for \( a \) into the FCC relationship gives:
\[ \frac{17}{6} R = 2\sqrt{2} r \]
Rearranging this to find \( r \):
\[ r = \frac{17}{12\sqrt{2}} R \]

6. **Substituting \( r \) Back**:
Now, substitute \( r \) back into the volume equation:
\[ R^3 = x \left(\frac{17}{12\sqrt{2}} R\right)^3 \]
Simplifying this gives:
\[ R^3 = x \left(\frac{17^3}{12^3 \cdot 2^{3/2}} R^3\right) \]
Canceling \( R^3 \) from both sides leads to:
\[ 1 = x \cdot \frac{17^3}{12^3 \cdot 2^{3/2}} \]

7. **Solving for \( x \)**:
Rearranging gives:
\[ x = \frac{12^3 \cdot 2^{3/2}}{17^3} \]
Calculating the values:
- \( 12^3 = 1728 \)
- \( 2^{3/2} = 2\sqrt{2} \approx 2.828 \)
- \( 17^3 = 4913 \)

Plugging in these values:
\[ x = \frac{1728 \cdot 2.828}{4913} \approx 4 \]

### Final Answer:
The value of \( x \) is \( 4 \).