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: 

1. $\frac{1}{2}\mathrm{a\; :}$ $\frac{\sqrt{3}}{4}\mathrm{a}$ $:$ $\frac{1}{{\displaystyle 2}\sqrt{2}}\mathrm{a}$

2. $\frac{1}{2}\mathrm{a\; :}$ $\sqrt{3}\mathrm{a}$ $:$ $\frac{1}{\sqrt{2}}\mathrm{a}$

3. $\frac{1}{2}\mathrm{a\; :}$ $\frac{\sqrt{3}}{2}\mathrm{a}$ $:$ $\frac{\sqrt{2}}{{\displaystyle 2}}\mathrm{a}$

4. $1\mathrm{a\; :}$ $\sqrt{3}\mathrm{a}$ $:$ $\sqrt{2}\mathrm{a}$

Silicon gives p-type of semiconductor on doping with:

1. Germanium.

2. Arsenic.

3. Selenium.

4. Boron.

For a cubic crystal structure, which one of the following relations indicating the cell characteristic is correct?

1. $a\ne b\ne c$ $and$ $\alpha \ne \beta$ $and$ $\gamma \ne 90°$

2. $a\ne b\ne c$ $and$ $\alpha =\beta$ $=$ $\gamma =90°$

3. $a=b=c$ $and$ $\alpha \ne \beta$ $=$ $\gamma =90°$

4. $a=b=c$ $and$ $\alpha =\beta =$ $\gamma =90°$

The fraction of total volume occupied by the atoms present in a simple cube is:

1. $\frac{\mathrm{\pi }}{6}$

2. $\frac{\mathrm{\pi }}{3\sqrt{2}}$

3. $\frac{\mathrm{\pi }}{4\sqrt{2}}$

4. $\frac{\mathrm{\pi }}{4}$

If NaCl is doped with 10^-4mol% of SrCl₂, the concentration of cation vacancies will be (N_A= 6.02×10²³mol^-1):

1. $6.02\times {10}^{15}$ $mo{l}^{-1}$

2. $6.02\times {10}^{16}$ $mo{l}^{-1}$

3. $6.02\times {10}^{17}$ $mo{l}^{-1}$

4. $6.02\times {10}^{14}$ $mo{l}^{-1}$

CsBr crystallises in a body-centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 amu and that of Br = 80 amu and the Avogadro number being 6.02×10²³ mol^-1, the density of CsBr is:

1. 42.5 $g/c{m}^3$

2. 0.425 $g/c{m}^3$

3. 8.25 $g/c{m}^3$

4. 4.25 $g/c{m}^3$

The appearance of colour in solid alkali metal halides is generally due to:

1. F-centres.

2. Schottky defect.

3. Frenkel defect.

4. Interstitial positions.

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75 % of octahedral voids. The formula of the compound is:

1. ${\mathrm{C}}_3{\mathrm{A}}_4$

2. ${\mathrm{C}}_2{\mathrm{A}}_3$

3. ${\mathrm{C}}_3{\mathrm{A}}_2$

4. ${\mathrm{C}}_4{\mathrm{A}}_3$