The Solid State

Question 261
CBSEENCH12010304

Aluminum crystallizes in an fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal? 

Solution

Aluminum crystallizes in FCC structure atomic radius (r) = 125 pm

Length of the side of the unit cell in FCC structure,  straight a equals fraction numerator 4 over denominator square root of 2 end fraction straight r space equals space 2 square root of 2 straight r end root

 Therefore            a= 2 x 1.414 x 125

a = 353.5 pm

Question 262
CBSEENCH12010324

What are n-types semiconductors?

Solution

The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. These are generated when the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As.

Question 263
CBSEENCH12010343

Copper crystallises with face centred cubic unit cell. If the radius of a copper atom is 127.8 pm, calculate the density of the copper metal.

(Atomic mass of Cu = 63.55 u and Avogadro’s number NA = 6.02 x 1023 mol-1)

Solution

No. of atoms in unit cell, Z = 4

Radius = 127.8 pm
Error converting from MathML to accessible text.

Question 264
CBSEENCH12010344

Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. The density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u)

Solution

a = 286.65 pm

= 286.65 x 10-10 cm

Density (Ï) = 7.874 g cm-3

At mass of Fe = 56.0 u

Z = 2 (For body centered cubic unit cell)

Avogadro number (N0) =?
Error converting from MathML to accessible text.

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