1

### AIPMT 2008

If $a$ stands for the edge length of the cubic systems: simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively
A
${1 \over 2}a:{{\sqrt 3 } \over 2}a:{{\sqrt 2 } \over 2}a$
B
$1a:\sqrt 3 a:\sqrt 2 a$
C
${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$
D
${1 \over 2}a:\sqrt 3 a:{1 \over {\sqrt 2 }}a$

## Explanation

For Simple cubic : r+ + r = ${a \over 2}$

For Body centred : r+ + r = ${{a\sqrt 2 } \over 4}$

For Face centered: r+ + r = ${a \over {2\sqrt 2 }}$

$\therefore$ Ratio of radii of the three will be

= ${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$
2

### AIPMT 2008

Which of the following statements is not correct?
A
The number of carbon atoms in a unit cell of diamond is 4
B
The number of Bravais lattices in which a crystal can be categorized is 14
C
The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
D
Molecular solids are generally volatile.

## Explanation

Packing fraction for a cubic unit cell is given by

f = ${{Z \times {4 \over 3}\pi {r^3}} \over {{a^3}}}$

where a = edge length, r = radius of cation and anion.

Efficiency of packing in simple cubic or primitive cell = π/6 = 0.52 or 52 %

It means 52 % of unit cell is occupied by atoms and 48 % is empty.
3

### AIPMT 2008

With which one of the following elements silicon should be doped so as to give p-type of semiconductor?
A
Selenium
B
Boron
C
Germanium
D
Arsenic

## Explanation

The semiconductors formed by the introduction of impurity atoms containing one elecron less than the parent atoms of insulators are termed as p-type semiconductors. Therefore silicon containing 14 electrons has to be doped with boron containing 13 electrons to give a p-type semi-conductor.
4

### AIPMT 2007

If NaCl is doped with 10$-$4 mol% of SrCl2, the concentration of cation vacancies will be (NA = 6.02 $\times$ 1023 mol$-$1)
A
6.02 $\times$ 1016 mol$-$1
B
6.02 $\times$ 1017 mol$-$1
C
6.02 $\times$ 1014 mol$-$1
D
6.02 $\times$ 1015 mol$-$1

## Explanation

As each Sr2+ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl2 added.

$\therefore$ Concentration of cation vacancies

= 10–4 mole %

= ${{{{10}^{ - 4}}} \over {100}} \times 6.023 \times {10^{23}}$

= $6.023 \times {10^{23}}$ $\times$ 10-6

= $6.023 \times {10^{17}}$