Explain why does conductivity of germanium crystals increase on doping with galium?
On doping germanium with galium some of the positions of lattice of germanium are occupied be galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied and the place remains vacant. This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position. Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.
In a compound, nitrogen atoms $(\mathrm{N})$ make cubic close packed lattic and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N ?
Let the number of N atoms in ccp is $x$
$$\begin{array}{lrl} \therefore & \text { Number of tetrahedral voids } & =2 x \\ \therefore & \text { Number of } M \text { atoms }=\frac{1}{3} \times 2 x \\ & \frac{\text { Number of } N \text { atoms }}{\text { Number of } M \text { atoms }}=\frac{3 x}{2 x}=\frac{3}{2} \end{array}$$
So, the formula of the compound is $M_2 N_3$.
Under which situations can an amorphous substance change to crystalline form?
On heating, amorphous substances change to crystalline form at some temperature some objects from ancient civilisation are found to be milky in appearance. This is due to crystallisation.
Match the defects given in Column I with the statements in given Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Simple vacancy defect | 1. | Shown by non-ionic solids and increases density of the solid |
| B. | Simple interstitial defect | 2. | Shown by ionic solids and decreases density of the solid |
| C. | Frenkel defect | 3. | Shown by non-ionic solids and density of the solid decreases |
| D. | Schottky defect | 4. | Shown by ionic solids and density of the solid remains the same |
A. $\rightarrow$ (3) B. $\rightarrow$ (1) C. $\rightarrow$ (4) D. $\rightarrow$ (2)
A. When some of lattice sites are vacant in any non-ionic solid, the crystal is said to have vacancy defect and due to decrease in number of particles present in crystal lattice the density of crystal decreases.
B. Simple interstitial defect are shown by non-ionic solids in which constituent particles is displaced from its normal site to an interstitial site. Hence, density of solid increases.
C. Frenkel defect is shown by ionic solids in which smaller ions get dislocated from its normal site to its interstitial site which lead to decrease its density.
D. Schottky defect is shown by ionic solids in which equal number of cation and anion get missed from ionic solids and thus, density of solid decreases.
Match the type of unit cell given in Column I with the features given in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Primitive cubic unit cell | 1. | Each of three perpendicular edges compulsorily have the different edge length i.e., $a\ne b \ne c$ |
| B. | Body centred cubic unit cell | 2. | Number of atoms per unit cell is one |
| C. | Face centred cubic unit cell | 3. | Each of the three perpendicular edges compulsorily have the same edge length i.e., $a=b=c$ |
| D. | End centred orthorhombic unit cell | 4. | In addition to the contribution from the corner atoms the number of atoms present in a unit cell is one |
| 5. | In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three |
A. $\rightarrow(2,3)$ B. $\rightarrow(3,4)$ C. $\rightarrow(3,5)$ D. $\rightarrow(1,4)$
A. For primitive unit cell, $a=b=c$
Total number of atoms per unit cell $=1 / 8 \times 8=1$
Here, $1 / 8$ is due to contribution of each atom present at corner.
B. For body centred cubic unit cell, $a=b=c$
This lattice contain atoms at corner as well as body centre. Contribution due to atoms at corner $=1 / 8 \times 8=1$ contribution due to atoms at body centre $=8$
C. For face centred unit cell, $a=b=c$
Total constituent ions per unit cell present at corners $=\frac{1}{8} \times 8=1$
Total constituent ions per unit cell present at face centre $=\frac{1}{2} \times 6=3$
D. For end centered orthorhombic unit cell, $a \neq b \neq c$
Total contribution of atoms present at corner $=\frac{1}{8} \times 8=1$
Total contribution of atoms present at end centre $=\frac{1}{2} \times 2=1$
Hence, other than corner it contain total one atom per unit cell.
