Question

[2×4/3πr3 / 4/√3r × 100 ] % (percentage) .
please solve this problem ( answer is 68% percent).​

Answer 1

Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

As sphere are touching each other

Therefore a = 2r

No. of spheres per unit cell = 1/8 × 8 = 1  

Volume of the sphere = 4/3 πr3  

Volume of the cube = a3= (2r)3 = 8r3

∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

∴ % occupied = 52.4 %

Answer 2

Answer:

36.75 ℅

Explanation:

.

[2×(4/3πr3 / 4/√3r)×100 ] %

=[ 2×(1/√3πr²)×100 ] %

=[ 2×(0.1837763)×100 ] %

=[ 0.367552599 ×100 ] %

=(36.75526) %

≈ 36.76 ℅