Q. 363.7( 3 Votes )

# The edge lengths of the unit cells in terms of the radius of spheres constituting fcc, bcc and simple cubic unit cell are respectively________.

A.

B.

C.

D.

Answer :

Referring explanation to Question 33, edge lengths can be calculated in terms of the radius of the spheres.

In case of fcc structure, since the atoms are located at the corners and the center of the face of the cube, the edge length can be calculated from the diagonal formed by joining the radii of the three spheres.

In ΔABC,

AC^{2} = b^{2} = BC^{2} + AB^{2}

= a^{2} + a^{2} = 2a^{2}

b = a

If *r* is the radius of the sphere, then b = 4r =

Or,

In case of bcc structure, the atoms are placed at the corners of the cube with one atom in the center of the cube. The central atom will be in touch with the other two diagonally arranged atoms.

In ΔEFD, FD^{2} = EF^{2} + ED^{2}

EF = ED = *a*, which is the side of the cube. FD = *b*.

b = a

In ΔAFD, let the diagonal AF = c.

c^{2} = a^{2} + b^{2} = a^{2} + 2a^{2} = 3a^{2}.

c = a.

The length of the diagonal of the cube is equal to 4 times of the radius of the sphere as the three spheres are along the diagonal.

Therefore, a = 4*r*

a = .

In case of simple cubic cell, the atoms are located only at the corners of the cubic unit cell with the spheres touching the edges of the adjacent spheres. Thus the edge length is equal to twice the radius of the sphere, 2*r*.

The correct option is (i).

Rate this question :

An element ‘*X*’ (At. mass = 40 g mol^{–1}) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of ‘*X*’ and the number of unit cells in 4 g of ‘*X*’. (*N _{A}* = 6.022 × 10

^{23}mol

^{–1})

Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

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NCERT - Chemistry Part-I