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a
A - 1 ; B - 2 ; C - 3
b
A - 3 ; B - 1 ; C - 2
c
A - 3 ; B - 2 ; C - 1
d
A - 2 ; B - 3; C - 1
answer is B.
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Detailed Solution
A→3;B→1;C→2[∵2 is simple cubic,3 is fcc and 1 is bcc]. (1) For bcc:In body centered cubic unit cell, one atom is located at body center apart from corners of the cube. Let us take a unit cell of edge length “a”. Length of body diagonal, c can be calculated with help of Pythagoras theorem,In ∆AFD, c2 = a2 + b2Where b is the length of face diagonal, thus b = 2 a c2 = 3a2 ; c = 3 aFrom the figure, radius of the sphere, r = 1/4 × length of body diagonal, c ⇒r=c4 = 34 a ; a = 43 r(2) For simple cubic lattice:In simple cubic unit cell, atoms are located at the corners of cube. Let us take a unit cell of edge length “a”. Radius of atom can be given as,r = a2 ; a = 2r(3) For ccp :Hexagonal close packing (hcp) and cubic close packing (ccp) have same packing efficiency. Let us take a unit cell of edge length “a”. Length of face diagonal, b can be calculated with the help of Pythagoras theorem,In ∆ABC, b2=a2+a2The radius of the sphere is rThe face diagonal (b) = r + 2r + r = 4r∴(4r)2=a2+a2⇒(4r)2=2a2⇒a=16r22 = 22r