Suppose a pure Si crystal has 5×10^28 atoms . . .

Question : Suppose a pure Si crystal has 5×10²⁸ atoms m^-3. It is doped by 1 ppm concentration of boron. Calculate the concentration of holes and electrons, given that n_i=1.5×10¹⁶ m^-3. Is the doped crystal n-type or p-type? 

Doubt by Nevaeh

Solution : 

Number of silicon atoms = 5×10²⁸ m^-3

Doping concentration = 1 ppm (parts per million) of Boron

Boron (B) is a trivalent impurity, it will accept one electron per atom.

so the hole concentration would be 

n_h=[5×10²⁸/10⁶] = 5×10²² m^-3
n_h= 5×10²² m^-3

Now, using the mass action law

n_i²=n_e.n_h
n_e=n_i²/n_h
n_e=[1.5×10¹⁶]²/[5×10²²]
n_e=(2.25×10³²)/[5×10²²]
n_e=0.45×10¹⁰
n_e=4.5×10⁹ m^-3

Clearly, n_h>n_e
Hence, doped crystal is a p-type semiconductor.