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If '$a$' stands for the edge length of the cubic systems : simple cubic, body centered cubic and face centered cubic, then the ratio of radii of the spheres in these systems will be respectively.
A
$\dfrac{1}{2}a : \dfrac{\sqrt{3}}{2}a : \dfrac{\sqrt{2}}{2}a$
B
$1a : \sqrt{3}a : \sqrt{2}a$
C
$\dfrac{1}{2}a : \dfrac{\sqrt{3}}{4}a : \dfrac{1}{2\sqrt{2}}a$
D
$\dfrac{1}{2}a : \sqrt{3}a : \dfrac{1}{\sqrt{2}}a$
Detailed Solution
For simple cubic: the spheres touch along the edge, so $r^+ + r^- = \dfrac{a}{2}$, i.e. $2r = a$ and $r = \dfrac{a}{2}$.
Here $a$ = edge length and $r^+ + r^-$ = interatomic distance.
For body centered cubic: the spheres touch along the body diagonal, so $4r = \sqrt{3}\,a$ and $r = \dfrac{\sqrt{3}\,a}{4}$.
For face centered cubic: the spheres touch along the face diagonal, so $4r = \sqrt{2}\,a$ and $r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}}$.
Therefore the ratio of radii of the three will be:
$\dfrac{a}{2} : \dfrac{\sqrt{3}\,a}{4} : \dfrac{a}{2\sqrt{2}}$
Here $a$ = edge length and $r^+ + r^-$ = interatomic distance.
For body centered cubic: the spheres touch along the body diagonal, so $4r = \sqrt{3}\,a$ and $r = \dfrac{\sqrt{3}\,a}{4}$.
For face centered cubic: the spheres touch along the face diagonal, so $4r = \sqrt{2}\,a$ and $r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}}$.
Therefore the ratio of radii of the three will be:
$\dfrac{a}{2} : \dfrac{\sqrt{3}\,a}{4} : \dfrac{a}{2\sqrt{2}}$