Question

Hi everyone, I have a problem about semiconductor.

I attempted once, but failed.

So please help me with it, thank you.

Simple Cubic Lattice, Part 2 For the simple cubic crystal described above, i.e. a = 0.7 nm, calculate the surface density of atoms (i.e. number of atoms per unit area) on the (100) plane in unit of cm 2. Values within 5% error will be considered correct. 408000000000000 ! Incorrect

Answer 1

The area of the simple cubic crystal of the (100) plane within the unit cell is :

For the (100) plane, there are 4 atoms at the 4 corners of the square and one atom in the middle.

One fourth of each corner atom is enclosed within the unit cell, and middle atom is entirely within the unit cell, so the number of atoms on the (100) plane within the unit cell is

$N_{100}= 4*\frac{1}{4} + 1*1$

Therefore,

Atomic density on the (100) plane =

$\eta =\frac{N_{100}}{A_{100}}$

$\eta =\frac{2}{a^2}$

$\eta =\frac{2}{(0.7*10^{-9})^2}$

$\eta =\frac{2}{(0.49*10^{-18})}$

$\eta =4.0816*10^{18}\frac{atoms}{cm^2}$

The answer has issue with the number of zeroes.

Yours is coming 14. However it is 18.