InAs is a lllV semiconductor that has the zinc blende struct

InAs is a lll-V semiconductor that has the zinc blende structure with a lattice parameter of 0.606 nm. Given the atomic masses of In (114.82 g mol-1) and As (74.92 g mol-1) find the density. CdO has the NaCI crystal structure with a lattice parameter of 0.4695 nm. Given the atomic masses of Cd (112.41 g mol-1) and O (16.00 g mol-1) find the density. KCI has the same crystal structure as NaCI. The lattice parameter a of KCI is 0.629 nm. The atomic masses of K and Cl are 39.10 g mol-1 and 35.45 g mol-1 respectively. Calculate the density of KCI.

Solution

(a) atomic mass of In ans As are 114.82g/mol and 74.92 g/mol respectively

114.82/6.02x10²³ gram/atom and 74.92/6.02x10²³ gram/atom respectively

In As is FCC with four atoms per unit cell

so the mass of In As if the sum of mass of 4 atoms of In and 4 atoms of As

mass of unit cell of InAs= (76.29x10^-23 +49.78x10-²³ )gram =126.07x10^-23 gram

volume of unit cell= cube of lattice parameter=(0.606x10^-9 )³ =222.54x10^{-30 m3}

thus density= mass/volume= 0.566x10⁷ gram/m³ = 5.67gram/cm³

(b) CdO is FCC with four atoms per unit cell

atomic masses of Cd and O are=112.41g/mol and 16 g/mol respectively

=112.41/6.02x10²³ g/atom and 16/6.02x10²³ g/atom

mass of 4 atoms of Cd and 4 atoms of O are= 74.69x10^{-23 gram} and 10.63x10^-23 gram respectively

mass of unit cell of CdO= (74.69+10.63)x10^-23 =85.32x10^{-23 gram}

volume of unit cell is cube of lattice parameter= (0.4695x10^-9 )³ =0.1034x10^-27 m³

thus density= mass/volume = 85.32x10^{-23 gram} /0.1034x10^-27 =825.14x10⁴ g/m3 =8.25gram/cm³

(c) KCl has FCC structure

it has 4 atoms per unit cell.

lattice parameter for KCl is=0.629 nm

atomic masses of K and Cl are= 39.10g/mo and 35.45g/mol respectively

                                                        =39.10/6.02x10²³ g/atom and 35.45/6.02x10²³ g/atom respectively

so the mass of 4 atoms of K and 4 atoms of cl are=25.98x10^-23 gram and 23.55x10^-23 gram

thus the mass of unit cell of KCl=(25.98+23.55)x10^-23 gram =49.53x10^{-23 gram}

volume of unit cell is cube of lattice parameter= (.629x10^-9 )³ =2.48x10^-28 m³

thus the density of KCl is =49.53x10^-23 /2.48x10^-28 = 19.97x10⁵ gram/m³ =1.997gram/cm³