Empty space present in FCC is

26%

The correct answer is option 2. 26%.

Let \(a\) be the edge length of the given unit cell.

Then from \(\Delta AC\), applying Pythagoras theorem,

\((AC)^2 = (AD)^2 + (DC)^2\)

or, \(b^2 = a^2 + a^2\)

Let \(r\) be the radius of the sphere, then

the length of the face diagonal, \((b) = r + 2r + r = 4r\)

Applying the value in the above equation, we get,

\((4r)^2 = a^2 + a^2\)

or, \(16r^2 = 2a^2\)

or, \(a^2 = \frac{16}{2}r^2\)

or, \(a = \sqrt{8r^2}\)

or, \(a = 2\sqrt{2}r\)

Therefore, volume of the cube is

\(a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3\)

We know volume of the sphere \(= \frac{4}{3}\pi r^3\)

Total number of particles in a fcc unit cell \(= 4\)

Therefore,

\(\text{Packing fraction = }\frac{\text{Total volume of the sphere}}{\text{Volume of the cube}} \times  100\)

or, \(\text{Packing fraction = }\frac{4 \times \frac{4}{3}\pi r^3}{16\sqrt{2}r^3} \times 100\)

or, \(\text{Packing fraction = } \frac{\frac{16}{3}\pi }{16\sqrt{2}r^3} \times 100\)

or, \(\text{Packing fraction = } \frac{\pi }{3\sqrt{2}} \times 100\)

or, \(\text{Packing fraction } \approx 74\%\)

Thus, empty space in fcc unit cell

\(= (100 -74)\% = 26%\)