1

### 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.
2

### 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

= $6.023 \times {10^{17}}$
3

### AIPMT 2007

The fraction of total volume occupied by the atoms present in a simple cube is
A
${\pi \over {3\sqrt 2 }}$
B
${\pi \over {4\sqrt 2 }}$
C
${\pi \over 4}$
D
${\pi \over 6}$

## Explanation

The maximum properties of the available volume which may be filled by hard sphere in simple cubic arrangement is ${\pi \over 6}$ or 0.52.
4

### AIPMT 2006

CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 $\times$ 1023 mol$-$1, the density of CsBr is
A
4.25 g/cm3
B
42.5 g/cm3
C
0.425 g/cm3
D
8.25 g/cm3

## Explanation

Density of CsBr = ${{Z \times M} \over {V \times {N_A}}}$

= ${{1 \times 213} \over {{{\left( {436.6 \times {{10}^{ - 10}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}$

= 4.25 g/cm3